Work and energy for a book carried and lifted 1.5 m

[kinski_girl]

K. if anyone is able to help me with this problem, that would be great.

A student carries a 1.0 kg physics book 3.0 m across the room, then lifts it 1.5 m to put it on a shelf. The entire task takes 20 s.
a) what is the total work done on the book?
b) what is the change in potential energy of the book?
c) what is the power output of the student?
d) if the book falls off the shelf a distance of 2.5 m to the floor, how much kinetic energy will it have when it hits the floor?
e) what is the speed of the book as it hits the floor?

i was able to do part c, but the rest is driving me nuts! Please help!

 

[marlon]

Well you know that the potential energy associated with gravity is mgh where h is the vertical distance that anobject is rised.

The work is equal to F*r*cos(x) where F is the force, r the traveled distance and x is the angle between the force vector and the traveled distance. If both F and r are parallel then the work is just F*r...

The power is the work F per unit of time : F/t...

This should get you going

marlon

 

[thepatientmental]

Hi K.,

I would be happy to assist you with solving this physics problem involving work and energy. Let's break it down step by step.

a) To find the total work done on the book, we need to calculate the work done in each step of the task and then add them together. Work is calculated by multiplying the force applied by the distance moved in the direction of the force. In the first step, the book is moved horizontally, so the work done is W = F*d. We know the mass of the book (1.0 kg) and the distance it is moved (3.0 m), so we can calculate the force required using the formula F = m*a, where a is the acceleration due to gravity (9.8 m/s^2). Plugging in the values, we get F = 1.0 kg * 9.8 m/s^2 = 9.8 N. Therefore, the work done in the first step is W = 9.8 N * 3.0 m = 29.4 J. In the second step, the book is lifted vertically, so the work done is W = m*g*h, where g is the acceleration due to gravity and h is the height lifted (1.5 m). Plugging in the values, we get W = 1.0 kg * 9.8 m/s^2 * 1.5 m = 14.7 J. Therefore, the total work done on the book is 29.4 J + 14.7 J = 44.1 J.

b) The change in potential energy of an object is equal to the work done on the object. In this case, the work done in lifting the book is equal to the change in potential energy, so the change in potential energy is 14.7 J.

c) To calculate the power output of the student, we need to divide the work done by the time taken. In this case, the work done is 44.1 J and the time taken is 20 s. Therefore, the power output is P = W/t = 44.1 J/20 s = 2.205 watts.

d) When the book falls off the shelf, it will have a certain amount of kinetic energy due to its motion. We can calculate this using the formula KE = 1/2 * m * v^2, where m is the mass