1. While helping a friend move, you push a 11.5 kg dolly with a force of 11.0 N.
ID: 2003894 • Letter: 1
Question
1.
While helping a friend move, you push a 11.5 kg dolly with a force of 11.0 N. Assume there are no other forces acting on the dolly. If the dolly starts at rest, how far does it move in 3.50 s?
1 m
2.
At a lumberjack competition an axe with a mass of 1.25 kg is thrown at a target with a speed of 41 m/s. When it hits the target, it penetrates to a depth of 0.055 m.
(a) What was the average force exerted by the target on the axe?
1 N
(b) If the mass of the axe is halved, and the force exerted by the target on the axe remains the same, by what multiplicative factor does the penetration depth change?
2
It doubles. It increases by a factor of the square root of two. It does not change. It decreases by a factor of the square root of two. It halves.
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3
Your 1500 kg car pulls a 540 kg trailer away from a stop sign with an acceleration of 2.00 m/s2. (Note the answers should be given in kN.)
(a) What is the net force exerted by the car on the trailer?
1 kN
(b) What force does the trailer exert on the car?
2 kN
(c) What is the net force acting on the car?
3 kN
4.
Two boxes sit next to, and touching, each other on a frictionless horizontal surface. The lighter box has a mass of 5.31 kg and the heavier box has a mass of 7.29 kg.
(a) Find the contact force between these boxes when a horizontal force of 6.1 N is applied to the light one.
1 N
(b) If the 6.1 N force is applied to the heavy box instead, is the contact force between the boxes the same as, greater than, or less than the contact force in part (a)?
2
(c) Verify your answer to part (b) by calculating the contact force in this case.
3 N
5.
A child sits in a wagon and together they have a mass of 25 kg. The wagon is pulled with a constant speed by a handle that is at an angle of 22° above the horizontal. If the normal force exerted on the wagon is 210 N, what is the force F applied to the handle?
1 N
6.
An object acted on by three forces moves with constant velocity. One force acting on the object is in the positive x direction and has a magnitude of 6.7 N; a second force has a magnitude of 4.5 N and points in the negative y direction. Find the direction and magnitude of the third force acting on the object.
1 N
2° (counterclockwise from the +x axis)
7.
A chain is used to lift a 1200 kg car vertically upward with an acceleration of 0.90 m/s2. What is the tension in the chain?
1 N
8.
A 22.5 kg box (m1) rests on a table.
(a) What is the weight of the box?
1 N
What is the normal force acting on it?
2 N (upward)
(b) A 12.5 kg box (m2) is placed directly on top of the 22.5 kg box.
Determine the normal force that the table exerts on the 22.5 kg box.
3 N (upward)
Determine the normal force that the 22.5 kg box exerts on the 12.5 kg box.
4 N (upward)
9.
Your friend has a weight of 127 lb when standing on a scale. You take the scale to an elevator where you notice your friend's weight on the scale reads 108 lb. What are the direction and magnitude of the elevator's acceleration?
Direction 1
Magnitude 2 m/s2
10.
When you lift a package with a force of 82 N, the package accelerates upward at a rate of a. If, instead, you lift with a force of 93 N, the package now has an acceleration of 8 multiplied by a.
(a) Find the weight of the package.
1 N
(b) Find the acceleration a.
2 m/s2
11.
One 3.1 kg paint bucket is hanging by a massless cord from another 3.1 kg paint bucket, also hanging by a massless cord, as shown below.
(a) If the buckets are at rest, what is the tension in each cord?
(lower cord) 1 N
(upper cord) 2 N
(b) If the two buckets are pulled upward with an acceleration of 1.30 m/s2 by the upper cord, calculate the tension in each cord.
(lower cord) 3 N
(upper cord)5
12)
Your friend has a pair of fuzzy dice with a mass of 1.32 g hanging from his rear-view mirror.
(a) When he accelerates from a stoplight, the dice deflect backward toward the rear of the car hanging at an angle of 6.18° relative to the vertical. Find the tension in the string holding the dice.
N
(b) At what angle to the vertical will the tension in the string be twice the weight of the dice?
°
Explanation / Answer
1.
The mass of dolly, m = 11.5 kg
The force, F = 11 N
Initial velocity, u = 0
The time, t = 3.5 sec
We have a formula for force = mass * acceleration
from this, a = force/mass = 11/11.5 = 0.956 m/s^2
We have a formula for the distance moved as given below
S = ut + (1/2)at^2
= 0 + 0.5 * 0.956 * (3.5)^2 = 5.86 m
So the distance traveled by the dolly, s = 5.86 m
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