Asked by Julia
1. A 2.5 kg mass at rest on a horizontal surface is acted upon by an applied horizontal force of 150 N. A frictional force of 100 N acts on the mass. How much work is done on the mass after 5 m?
I got 250 J but I'm not sure if that's correct.
2. A mass of 10 kg, initially at rest on a horizontal frictionless surface, is acted upon by a horizontal force of 25 N. What is the speed of the mass after it has moved 5.0 m?
I got 5 m/s for this but apparently that's incorrect.
3. a) A jet engine applies a force of 300 N horizontally to the right against a 50 kg rocket. The force of air friction 200 N. If the rocket is thrust horizontally 2.0 m, the kinetic energy gained by the rocket is:
I'm not sure how to find the answer for this one.
3. b) A jet engine applies a force of 300 N horizontally to the right against a 50 kg rocket. The force of air friction 200 N. If the rocket is thrust horizontally 2.0 m, the velocity of the rocket is:
Not sure how to answer this one either.
How should I solve these to get the correct answers? Help with any of them would be great!
I got 250 J but I'm not sure if that's correct.
2. A mass of 10 kg, initially at rest on a horizontal frictionless surface, is acted upon by a horizontal force of 25 N. What is the speed of the mass after it has moved 5.0 m?
I got 5 m/s for this but apparently that's incorrect.
3. a) A jet engine applies a force of 300 N horizontally to the right against a 50 kg rocket. The force of air friction 200 N. If the rocket is thrust horizontally 2.0 m, the kinetic energy gained by the rocket is:
I'm not sure how to find the answer for this one.
3. b) A jet engine applies a force of 300 N horizontally to the right against a 50 kg rocket. The force of air friction 200 N. If the rocket is thrust horizontally 2.0 m, the velocity of the rocket is:
Not sure how to answer this one either.
How should I solve these to get the correct answers? Help with any of them would be great!
Answers
Answered by
R_scott
1. yes ... 250 J is correct
2. kinetic energy = work ... 1/2 * m * v^2 = f * d
... find the work , then solve for the velocity
... 5 m/s seems correct ... maybe a significant figure issue?
3. similar to #2
... work on the rocket is "net force" multiplied by distance
2. kinetic energy = work ... 1/2 * m * v^2 = f * d
... find the work , then solve for the velocity
... 5 m/s seems correct ... maybe a significant figure issue?
3. similar to #2
... work on the rocket is "net force" multiplied by distance
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