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Andru [333]
3 years ago
14

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
Physics
1 answer:
zubka84 [21]3 years ago
6 0

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

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Answer:

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Explanation:

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We have a quadratic equation, that, after simplifying terms, can be solved as follows, applying the quadratic formula:

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b) If we take as our new initial status the moment at which the spring is compressed, and the cookie is at rest, all the energy is potential:

E = Ui = 1/2 k d²

In this case, d is the same value that we got in a), i.e., 0.275 m (as the distance travelled by the cookie after going through the equilibrium point is the same length that the spring have been compressed).

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When the cookie passes again through the equilibrium position, the energy will be in part kinetic, and in part, it will have become thermal energy again.

So, we can write the following equation:

Kf = Ui - Ff.d = 16.6 J - 11.0 (0.275) m = 16.6 J - 3.03 J = 13.6 J

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