<p>A 70 kg person moves a mini fridge from the first floor to the scond floor of a house. The person pushes the 50 kg fridge up a ramp of length 6.1 m and a height of 4.05 m. He is able to move the fridge up the ramp at a constant speed.</p>
<p>a) How much work is done by the person in moving the fridge and himself to the second floor?</p>
<p>b) If it took the person 10.2 seconds to get the fridge and himself to the second floor, what was mechanical power output of the person?</p>
<p>I’m assuming that the ramp is an incline. With the incline being the hypotenuse, I’m assuming 6.1 and 4.05 are the “legs” of the triangle.</p>
<p>First draw yourself a free-body diagram of the box. Note that the axes should be perpendicular to the incline, in which, Mg would have a x and y component. The y component balances with the Normal force. The x component is what you are focusing on. Of importance is that the box is moving at constant speed. That means all forces have to balance. Thus, the sum of the forces in the x-direction have to be 0. In other words, the applied force has to equal the x-component of Mg.</p>
<p>From there use trig/pythagorean’s theorum to find theta/how long the incline is. Since the applied force is in the direction of motion, and since the force is constant (Mg and theta don’t change), Work simply equals F times distance the box is moved.</p>
<p>For part (B), power = work/time. Plug and chug.</p>
<p>a) okay so my work is as follows
W = Force x distance, in which each are parallel to each other
W = (1177.2N x sin(41.6)) x 6.1m
W = 4767.601 J
<em>I used 9.81 for the force of gravity</em></p>
<p>b) P = Work/Time
P = (4767.601 J / 10.2seconds)
P = 467.412 Watts</p>
<p>lol sry, we just learned about power today so part b makes more sense now</p>