PHYSICS MIDDLE SCHOOL

5) Consider pushing a 50.0 kg box through a 5.00 m displacement on both a flat surface and up aramp inclined to the horizontal by 15.0°. In both cases, you apply a force of 100.N parallel to the
surface (parallel to the floor or parallel to the ramp). Calculate the work done by:
a) the gravitational force as the box is pushed across the flat ground
b) the gravitational force as the box is pushed up the ramp
c) the force you apply as the box is pushed across the flat ground
d) the force you apply as the box is pushed up the ramp

Answers

Answer 1

Answer:

a) The work done by the gravitational force on the flat surface is zero

b) The work done by the gravitational force on the ramp is -634 J

c) The work done by the applied force on the flat surface is 500 J

d) The work done by the applied force on up along the ramp is 500 J

Explanation:

The work done by a force is given by the equation

$W=Fdcos \theta$

where

F is the magnitude of the force

d is the dispalcement of the object

$\theta$ is the angle between the direction of the force and of the displacement

In this problem, we want to calculate the work done by the gravitational force as the box is pushed across the flat ground.

We immediately notice that the gravitational force acts downward, while the displacement is horizontal: therefore, the angle between force and displacement is $90^(\circ)$ ; this means that $cos 90^(\circ)=0$ , and therefore, the work done is zero:

$W=0$

In this case, the box is pushed along the ramp. We have:

$F=mg=(50.0)(9.8)=490 N$ is the magnitude of the force of gravity, where

m = 50.0 kg is the mass of the box

$g=9.8 m/s^2$ is the acceleration of gravity

d = 5.00 m is the displacement of the box along the ramp

The ramp is inclined to the horizontal by $15.0^(\circ)$ , therefore the angle between the force of gravity and the displacement of the box (moving up along the ramp) is:

$\theta=90^(\circ)+15^(\circ)=105^(\circ)$

Therefore, the work done by gravity in this case is:

$W=(490)(5.00)(cos 105^(\circ))=-634 J$

In this case, we want to calculate the work done by the force you apply as the box is pushed across the flat ground.

Here we have:

F = 100.0 N (force applied)

d = 5.00 m (displacement of the box)

$\theta=0^(\circ)$ (the force is applied parallel to the flat surface, therefore force and displacement have same direction)

Therefore, the work done by the force you apply on the flat ground is:

$W=(100.0)(5.00)(cos 0^(\circ))=500 J$

In this last case, we want to calculate the work done by the force you apply as the box is pushed up along the ramp.

This time we have:

F = 100.0 N (force applied is the same)

d = 5.00 m (displacement of the box is also the same)

$\theta=0^(\circ)$ (the force is applied parallel to the ramp, therefore force and displacement have again same direction)

Therefore, the work done by the force you apply while pushing the box along the ramp is:

$W=(100.0)(5.00)(cos 0^(\circ))=500 J$

Learn more about work:

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