Question

You are unloading a refrigerator from a delivery van. The ramp on the van is 5.0 m long, and its top end is 1.4 m above the ground. As the refrigerator moves down the ramp, you are on the down side of the ramp trying to slow the motion by pushing horizontally against the refrigerator with a force of 270N .


Part A


How much work do you do on the refrigerator during its trip down the ramp?


W = ???

Answer 1

Answer:

W = 1296 J

Explanation:

First, we have to find the angle of the ramp as:

θ = $sin^{-1}$(1.4m/5m)

θ = 16.26°

Then, the work W is define as:

$W = F_xd$

where $F_x$ F is the force in the direction of the displacement and d the distance, so:

$F_x=F*cos(16.26)$

Then, replacing values, we get:

$W = (270cos(16.26))(5)$

W = 1296 J

Answer 2

The work is done for refrigerator during its trip down the ramp while unloading it from a delivery van is 1296 joules.

What is work done?

Work done is the force applied on a body to move it over a distance. Work done for inclined plane can be given as,

$W=F\cos(\theta)d$

Here (F) is the magnitude of force and (d) is the distance traveled.

The length of the ramp is 5 meter long and the top end of the ramp is 1.4 meter above the ground.
The image showing the ramp and van is attached below. The sine angle for the figure can be given as,

$\sin\theta=\dfrac{1.4}{5}\\\theta=\sin^{-1}(0.28)\\\theta=16.26^o$

Now the work done is the product of force into the displacement.

As the total distance the refrigerator has to travel is 5 meters and the force applied is 270 N. Thus the work done is,

$W=270\times\cos(16.26)\times5\\W=1296\rm J$

Thus the work is done for refrigerator during its trip down the ramp while unloading it from a delivery van is 1296 joules.

Learn more about the work done here;

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