Work question (physics)

Question

Question 18:A girl pushes a 20.0 kg box, initially at rest, up a frictionless

10.0m long ramp inclined 30.0° above the horizontal. If the

final velocity of the box at the top of the ramp is 1.00 m/s,

what was the work done on the box by the girl? (so, you need

to know the force that the girl applies)

Start by drawing the free-body diagram to determine the

forces acting on the box.

Determine, by Newton's 2nd Law, the applied force Fa. Here,

work is being done by the applied force FA to overcome the

force of gravity Fg and also accelerate the object with a net

force EF.

The answer is this : 991 J, I dont know how it was found though

Answer 1

Work = Force * distance (in the same direction) and

Work = change in KE

In this scenario, we are doing work by fighting against a force (the weight of the box) and we are also doing work by adding kinetic energy to the system to cause the speed of 1m/s at the top of the ramp. Let’s work with each of those separately and add our final work together at the end.

Work = force * distance (same direction)

The force that the girl is fighting is the Weight of the box

Weight box = mg = (20)(9.8) = 196N

The weight of the box pushes down, so the force required to counteract this weight and create a constant speed must push 196N upward

With F = 196 point upward, we can only use our vertical distance and NOT the length of the ramp because force and distance must point in the SAME direction in order to use the work equation.

But how tall is the ramp? Trig might help :)

Length of ramp = hypotenuse = 10 meters

Angle = 30 degrees

We want the height = opposite side

use sine = opp/hypotenuse

sin(30) = height/10

1/2 = height / 10

height of ramp = 5 meters tall

Distance = 5

Work done to fight the weight = F * d = 196*5 = 980N

Now we need the extra work that was done to cause the object to gain speed.

Work = change in kinetic energy

Change in KE = final KE - initial KE

KE = 1/2mv^2

Initial KE: the object started from rest, so initially it had no kinetic energy

initial KE = 0

Final KE: the object ended at 1 m/s

Final KE = 1/2 * 20 * 1^2 = 10 J

Work = change in KE

Work = 10 - 0 = 10J

An extra 10J of work was done to cause the change in kinetic energy.

Total work done = 980 + 10 = 990J

Answer 2

By Theorem of Change of Kinetic Energy work W = mv²/2 + mglsin30° = 20kg·1²m²/s²/2 + 20kg·9/81m/s²·10m·0.5 = 991 J