Question

A person uses a constant force to push a 14.0 kg crate 1.80 m up a frictionless 10⁰ incline and to also increase its speed from 0.5 m/s to 1.5 m/s in the process. Find: (a) the work done by gravity; (b) the change in kinetic energy of the crate; (c) the net work done on the crate; (d) the work done by the person

Answer 1

a) Making h as the shorter cathetus, we have,

$h=1.8*sin10=0.312$

We proceed to calculate the work done by gravity

$W_g=-mgh$

$W_g=-140*0.312=-43.76J$

b) It's necessary to calculate the Kinetic Energy, we use the equation of KE,

$\Delta KE = \frac{1}{2}m(v_2^2-v_1^2)$

$\Delta KE = \frac{1}{2}14(1.5^2-0.5^2)=14J$

c) By work energy theorem

$W_{net}=\Delta KE \rightarrow W_net = 14J$

d) we calculate net work through the law of conservation

$W_{net}=W_f+W_g=14$

$\rightarrow W_f = 14-W_g=14-(-43.76)$

$W_f=57.76J$