10. A 2-kg object is moving at 3m/s. A 4-N force is applied in the direction of motion and then
removed after the object has traveled an additional 5m. The work done by this force is:
A. 12 J
B. 15 J
C. 18 J
D. 20 J
E. 38 J
ans: D
11. A sledge (including load) weighs 5000 N. It is pulled on level snow by a dog team exerting a
horizontal force on it. The coefficient of kinetic friction between sledge and snow is 0.05. How
much work is done by the dog team pulling the sledge 1000m at constant speed?
A. 2.5 × 104 J
B. 2.5 × 105 J
C. 5.0 × 105 J
D. 2.5 × 106 J
E. 5.0 × 106 J
ans: B
12. Camping equipment weighing 6000N is pulled across a frozen lake by means of a horizontal
rope. The coefficient of kinetic friction is 0.05. The work done by the campers in pulling the
equipment 1000m at constant velocity is:
A. 3.1 × 104 J
B. 1.5 × 105 J
C. 3.0 × 105 J
D. 2.9 × 106 J
E. 6.0 × 106 J
ans: C
13. Camping equipment weighing 6000N is pulled across a frozen lake by means of a horizontal
rope. The coefficient of kinetic friction is 0.05. How much work is done by the campers in
pulling the equipment 1000m if its speed is increasing at the constant rate of 0.20m/s2?
A. −1.2 × 106 J
B. 1.8 × 105 J
C. 3.0 × 105 J
D. 4.2 × 105 J
E. 1.2 × 106 J
ans: D
14. A 1-kg block is lifted vertically 1m by a boy. The work done by the boy is about:
A. 1 ft · lb
B. 1 J
C. 10 J
D. 0.1J
E. zero
ans: C
88 Chapter 7: KINETIC ENERGY AND WORK
11. Ws = 5000 N. = Normal force(Fn).
Fk = u*Fn = 0.05 * 5000 = 250 N. = Eorce
of kinetic friction.
Fap-Fk = M*a
Fap - 250 = M*0 = 0
Fap = 250 N. = Force applied.
Work = Fap * d = 250 * 1000 = 250,000 J.
= 2.5*10^5 J.
12. Same procedure as #11.13.
13. Ws = 6000 N. = Normal force(Fn).
Fk = u*Fn = 0.05 * 6000 = 300 N.
M*g = 6000
M * 9.8 = 6000
M = 612 kg
Fap-Fk = M*a
Fap - 300 = 612*0.20
Solve for Fap.
Work = Fap * d
14. Wb = F = M*g = 1 * 10 = 10 N.
Work = F*d
To find the work done by a force, we can use the formula W = F * d * cos(theta), where W is the work done, F is the force applied, d is the distance traveled, and theta is the angle between the force and the direction of motion.
Let's go through each question step by step to find the work done:
10. The force applied is 4 N and the distance traveled is 5 m. Since the force is applied in the direction of motion, the angle between the force and the direction of motion is 0 degrees. Plugging these values into the formula, we get W = 4 * 5 * cos(0) = 4 * 5 * 1 = 20 J.
11. The force applied is the dog team exerting a horizontal force, and we need to find the work done in pulling the sledge. The coefficient of kinetic friction is given as 0.05. Since the sledge is pulled at a constant speed, the net force acting on it is zero. Therefore, the force applied by the dog team is equal in magnitude and opposite in direction to the force of friction. The force of friction is given by F_friction = friction coefficient * normal force. The normal force is equal to the weight of the sledge, which is given as 5000 N. Plugging in the values, we get F_friction = 0.05 * 5000 = 250 N. The distance traveled is 1000 m. Now, we can calculate the work done using the formula W = F * d * cos(theta), where theta is the angle between the force and the direction of motion. Since the force of friction is acting in the opposite direction of motion, the angle between the force of friction and the direction of motion is 180 degrees. Plugging in the values, we get W = (-250) * 1000 * cos(180) = -250 * 1000 * (-1) = 250000 J. Since the work done cannot be negative, we take the absolute value, and the answer is 250000 J = 2.5 * 10^5 J.
12. Similar to question 11, the force applied is the force to overcome friction. The coefficient of kinetic friction is given as 0.05. The normal force is equal to the weight of the equipment, which is given as 6000 N. Plugging in the values, we get F_friction = 0.05 * 6000 = 300 N. The distance traveled is 1000 m. Using the formula W = F * d * cos(theta), where the angle between the force of friction and the direction of motion is 180 degrees, we get W = (-300) * 1000 * cos(180) = 300000 J = 3 * 10^5 J.
13. The work done in this case involves both the force applied to overcome friction and the force applied to accelerate the equipment. The force applied to overcome friction is the same as in question 12, which is 300 N. The force applied to accelerate the equipment can be calculated using the formula F = m * a, where m is the mass and a is the acceleration. The mass of the equipment can be calculated using the weight and acceleration due to gravity, which is approximately 9.8 m/s^2. So, m = 6000 / 9.8 = 612.24 kg. The force applied to accelerate the equipment is F = 612.24 * 0.2 = 122.448 N. The total force applied is the vector sum of the force to overcome friction and the force to accelerate the equipment, which is approximately 300 + 122.448 = 422.448 N. The distance traveled is 1000 m. Using the formula W = F * d * cos(theta), where the angle between the total force applied and the direction of motion is 180 degrees, we get W = (-422.448) * 1000 * cos(180) = 422448 J = 4.2 * 10^5 J.
14. The force applied is the weight of the block, which is equal to the mass times acceleration due to gravity, which is approximately 1 kg * 9.8 m/s^2 = 9.8 N. The distance lifted is 1 m. Using the formula W = F * d * cos(theta), where theta is the angle between the force and the direction of motion, we get W = 9.8 * 1 * cos(0) = 9.8 J. Since the work done cannot be negative, the answer is 9.8 J.
So, the answers are:
10. D. 20 J
11. B. 2.5 × 10^5 J
12. C. 3.0 × 10^5 J
13. D. 4.2 × 10^5 J
14. C. 10 J
To find the work done in each of these scenarios, we can use the equation for work:
Work = Force * Distance * cos(theta)
where the force is the applied force, the distance is the distance traveled, and theta is the angle between the force and the direction of motion.
10. In this scenario, the force is applied in the direction of motion, so theta = 0 degrees. The force is 4 N, and the distance traveled after the force is removed is 5 m. Plugging these values into the equation, we get:
Work = 4 N * 5 m * cos(0) = 20 J
Therefore, the work done by the force is 20 J, so the answer is (D).
11. In this scenario, the force being exerted by the dog team is in the horizontal direction, and the sledge is also moving horizontally. This means that theta = 0 degrees. The weight of the sledge is 5000 N, and the distance traveled is 1000 m. The coefficient of kinetic friction is 0.05.
First, we need to find the frictional force acting on the sledge. The frictional force is given by:
Frictional force = coefficient of kinetic friction * weight
Frictional force = 0.05 * 5000 N = 250 N
Now, we can calculate the work done by the dog team:
Work = Force * Distance * cos(theta)
Work = (force of the dog team - frictional force) * distance * cos(0)
Work = (Force of the dog team - 250 N) * 1000 m * cos(0)
Since the sledge is being pulled at a constant speed, the force exerted by the dog team is equal to the frictional force:
Force of the dog team = 250 N
Plugging this value into the equation, we get:
Work = (250 N - 250 N) * 1000 m * cos(0) = 0 J
Therefore, the work done by the dog team is 0 J, so the answer is (E).
You can follow similar steps to find the work done in the other scenarios (12, 13, and 14) using the given information in each question.