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What is weight of a person who has a mass of 75 kg?
736 N
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | W = mg | Weight (W) is the product of mass (m) and the acceleration due to gravity (g). |
| 2 | W = 75, \text{kg} \times 9.81, \text{m/s}^2 | Substituting m = 75, \text{kg} and g = 9.81, \text{m/s}^2. |
| 3 | \boxed{735.75, \text{N}} (Newtons) | Final Answer |
What force would have to be applied to start a 12.3 kg wood block moving on a surface with a static coefficient of friction of 0.438?
A 1509 g wood block is being pulled by the force meter at a constant velocity. Using the graph below find:
Two blocks made of different materials, connected by a thin cord, slide down a plane ramp inclined at an angle \theta to the horizontal. If the coefficients of friction are µA = .2 and µB = .3 and if mA = mB = 5.0 kg determine
A force F is used to hold a block of mass m on an incline as shown in the diagram above. The plane makes an angle of \theta with the horizontal and F is perpendicular to the plane. The coefficient of friction between the plane and the block is µ. What is the minimum force, F, necessary to keep the block at rest?
When the speed of a rear-wheel-drive car is increasing on a horizontal road, what is the direction of the frictional force on the tires?
Block m1 is stacked on top of block m2. Block m2 is connected by a light cord to block m3, which is pulled along a frictionless surface with a force F as shown in the diagram above. Block m1 is accelerated at the same rate as block m2 because of the frictional forces between the two blocks. If all three blocks have the same mass m, what is the minimum coefficient of static friction between block m1 and block m2?
In the diagram shown a 20 N force is applied to a block B (7 kg). Block A has a mass of 3 kg. Assume frictionless conditions.
In the diagram below, A has a mass of 3.2 kg and B a mass of 2.4 kg. The pulley is frictionless and has no mass.
Three identical blocks are being pulled or pushed across a rough horizontal surface by force of identical magnitude F, as shown in the drawing below. Rank the kinetic frictional forces that act on the blocks from smallest to greatest.
Traveling at a speed of 15.9 m/s, the driver of an automobile suddenly locks the wheels by slamming on the brakes. The coefficient of kinetic friction between the tires and the road is 0.659. What is the speed of the automobile after 1.59 s have elapsed? Ignore the effects of air resistance.
A person is trying to judge whether a picture (mass = 1.42 kg) is properly positioned by temporarily pressing it against a wall. The pressing force is perpendicular to the wall. The coefficient of static friction between the picture and the wall is 0.62. What is the minimum amount of pressing force that must be used?
A space probe far from the Earth is traveling at 14.8 km/s. It has mass 1312 kg. The probe fires its rockets to give a constant thrust of 156 kN for 220 seconds. It accelerates in the same direction as its initial velocity. In this time it burns 150 kg of fuel. Calculate final speed of the space probe in km/s.
Note: This is a bonus question. Skip if you haven’t yet taken calculus.
The alarm at a fire station rings and a 79.34-kg fireman, starting from rest, slides down a pole to the floor below (a distance of 4.20 m). Just before landing, his speed is 1.36 m/s. What is the magnitude of the kinetic frictional force exerted on the fireman as he slides down the pole?
A person whose weight is 4.92 × 102 N is being pulled up vertically by a rope from the bottom of a cave that is 35.2 m deep. The maximum tension that the rope can withstand without breaking is 592 N. What is the shortest time, starting from rest, in which the person can be brought out of the cave?
Two objects (49.0 and 24.0 kg) are connected by a massless string that passes over a massless, frictionless pulley. The pulley hangs from the ceiling. Find the acceleration of the objects and the tension in the string.
A train consists of 50 cars, each of which has a mass of 6.1 x 103 kg. The train has an acceleration of 8.0 × 10-2 m/s2?. Ignore friction and determine the tension in the coupling at the following places:
A block rests on a flat plane inclined at an angle of 30° with respect to the horizontal. What is the minimum coefficient of friction necessary to keep the block from sliding?
A student is watching their hockey puck slide up and down an incline. They give the puck a quick push along a frictionless table, and it slides up a 30° rough incline (µk = .4) of distance d, with an initial speed of 5 m/s, and then it slides back down.
Does it take longer to move up the distance d or back down the distance d? Or does it take the same amount of time?
Bonus Challenge: Repeat the problem but assume you are not given the initial speed, angle of incline, or µk.
A 25.0-kg box is released on a 23.5° incline and accelerates down the incline at 0.35 m/s2. Find the friction force impeding its motion. What is the coefficient of kinetic friction?
The coefficient of static friction between hard rubber and normal street pavement is about 0.85. On how steep a hill (maximum angle) can you leave a car parked?
When a basketball is dropped to the pavement, it bounces back up. Is a force needed to make it bounce back up? If so, what exerts the force?
An object has a mass of 10 kg. For each case below answer the questions and provide an example.
A 1.5 kg block is pushed to the right with just enough force to get it to move. The block is pushed for five seconds with this constant force, then the force is released and the block slides to a stop. If the coefficient of kinetic friction is 0.300 and the coefficient of static friction is 0.400. Calculate the amount of time that passes from when the force is applied to when the block stops.
A 2.0 kg wood box slides down a vertical wood wall while you push on it at a 45 ° angle. The coefficient of kinetic friction of wood µk = 0.200. What magnitude of force should you apply to cause the box to slide down at a constant speed?
A rocket-powered hockey puck has a thrust of 4.40 N and a total mass of 1.00 kg . It is released from rest on a frictionless table, 2.10 m from the edge of a 2.10 m drop. The front of the rocket is pointed directly toward the edge. Assuming that the thrust of the rocket present for the entire time of travel, how far does the puck land from the base of the table?
A car slides up a frictionless inclined plane. How does the normal force of the incline on the car compare with the weight of the car?
Two students push a 1750 kg car with a force of 758 N along a perfectly level road at a constant velocity of 4.00 m/s. Find the force of friction.
A 45 kg crate accelerates at 1.65 m/s2 when pulled by a rope with a force of 200 N. Find the angle the rope is pulled at. Friction is negligible.
A 135.0 N force is applied to a 30.0 kg box at 42 degree angle to the horizontal. If the force of friction is 85.0, what is the net force and acceleration? If the object starts from rest, how far has it traveled in 3.3 sec?
A skydiver reaches a terminal velocity of 55.0 m/s. At terminal velocity, the skydiver no longer accelerates. The mass of the skydiver and her equipment is 87.0 kg. What is the force of friction acting on her?
A small sphere hangs from a string attached to the ceiling of a uniformly accelerating train car. It is observed that the string makes an angle of 37° with respect to the vertical. The magnitude of the acceleration a of the train car is most nearly:
What is the mass of a dog that weighs 58 N on Earth?
A 1100 kg car accelerates from 32 m/s to 8.0 m/s in 4.0 sec. What amount of force was needed to slow it down?
An object weighs 300 N on Earth and 50 N on the moon. Does the object have less inertia on the moon?
Which of the following statements about the acceleration due to gravity is TRUE?
736 N
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| Kinematics | Forces |
|---|---|
| \Delta x = v_i \cdot t + \frac{1}{2} a \cdot t^2 | F = m \cdot a |
| v = v_i + a \cdot t | F_g = \frac{G \cdot m_1 \cdot m_2}{r^2} |
| a = \frac{\Delta v}{\Delta t} | f = \mu \cdot N |
| R = \frac{v_i^2 \cdot \sin(2\theta)}{g} |
| Circular Motion | Energy |
|---|---|
| F_c = \frac{m \cdot v^2}{r} | KE = \frac{1}{2} m \cdot v^2 |
| a_c = \frac{v^2}{r} | PE = m \cdot g \cdot h |
| KE_i + PE_i = KE_f + PE_f |
| Momentum | Torque and Rotations |
|---|---|
| p = m \cdot v | \tau = r \cdot F \cdot \sin(\theta) |
| J = \Delta p | I = \sum m \cdot r^2 |
| p_i = p_f | L = I \cdot \omega |
| Simple Harmonic Motion |
|---|
| F = -k \cdot x |
| T = 2\pi \sqrt{\frac{l}{g}} |
| T = 2\pi \sqrt{\frac{m}{k}} |
| Constant | Description |
|---|---|
| g | Acceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface |
| G | Universal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2 |
| \mu_k and \mu_s | Coefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion. |
| k | Spring constant, in \text{N/m} |
| Variable | SI Unit |
|---|---|
| s (Displacement) | \text{meters (m)} |
| v (Velocity) | \text{meters per second (m/s)} |
| a (Acceleration) | \text{meters per second squared (m/s}^2\text{)} |
| t (Time) | \text{seconds (s)} |
| m (Mass) | \text{kilograms (kg)} |
| Variable | Derived SI Unit |
|---|---|
| F (Force) | \text{newtons (N)} |
| E, PE, KE (Energy, Potential Energy, Kinetic Energy) | \text{joules (J)} |
| P (Power) | \text{watts (W)} |
| p (Momentum) | \text{kilogram meters per second (kg·m/s)} |
| \omega (Angular Velocity) | \text{radians per second (rad/s)} |
| \tau (Torque) | \text{newton meters (N·m)} |
| I (Moment of Inertia) | \text{kilogram meter squared (kg·m}^2\text{)} |
| f (Frequency) | \text{hertz (Hz)} |
General Metric Conversion Chart
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: \text{5 km}
Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}
Perform the multiplication: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}
Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | 10^{-12} | |
Nano- | n | 10^{-9} | |
Micro- | µ | 10^{-6} | |
Milli- | m | 10^{-3} | |
Centi- | c | 10^{-2} | |
Deci- | d | 10^{-1} | |
(Base unit) | – | 10^{0} | |
Deca- or Deka- | da | 10^{1} | |
Hecto- | h | 10^{2} | |
Kilo- | k | 10^{3} | |
Mega- | M | 10^{6} | |
Giga- | G | 10^{9} | |
Tera- | T | 10^{12} |
