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Allisa [31]
3 years ago
11

2.10-kg box is moving to the right with speed 8.50 m>s on a horizontal, frictionless surface. At t = 0 a horizontal force is

applied to the box. The force is directed to the left and has magnitude F1t2 = 16.00 N>s22t2. (a) What distance does the box move from its position at t = 0 before its speed is reduced to zero? (b) If the force continues to be applied, what is the speed of the box at t = 3.00 s?

Physics
2 answers:
larisa86 [58]3 years ago
8 0

Explanation:

Below is an attachment containing the solution.

vodka [1.7K]3 years ago
4 0

Answer:

a) the distance moved by the object is 14m

b) the speed of the box at 3.00s is - 18.5m/s

The negative sign in the answer indicates that the speed of the box is in opposite direction. Therefore, the direction of the box is towards left.

Explanation:

see the attached file

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Answer:

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Before the explosion we have p_i=m_iv_i=Mv_i.

After the explosion we have pieces 1 and 2, so p_f=m_1v_1+m_2v_2.

These equations must be vectorial but since we look at the instants before and after the explosions and the bomb fragments in only 2 pieces the problem can be simplified in one dimension with direction east-west.

Since we know momentum must be conserved we have:

Mv_i=m_1v_1+m_2v_2

Which means (since we want v_2 and M=m_1+m_2):

v_2=\frac{Mv_i-m_1v_1}{m_2}=\frac{Mv_i-m_1v_1}{M-m_1}

So for our values we have:

v_2=\frac{(145kg)(24m/s)-(115kg)(65m/s)}{(145kg-115kg)}=-133.17m/s

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