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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

User CMW
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1 Answer

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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 = (1)/(2)m(v_2^2-v_1^2)


\Delta KE = (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

User Chrstpsln
by
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