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Paha777 [63]
3 years ago
13

A block slides forward 4.73 m while being pulled back by a 39.7 N force at 142° from the direction of motion. How much work does

the force do?
Physics
1 answer:
lubasha [3.4K]3 years ago
6 0

Answer:

work done by the force, W = -147.97 J

Given:

displacement of the box, s = 4.73 m

angle between force and displacement, \theta = 142^{\circ}

Force, F = 39.7 N

Solution:

Work done by the force can be expressed as the dot product of Force, F and displacement 's' is given by:

W = Fscos\theta

W = 39.7\times 4.73\times cos142^{\circ}

W = -147.97 J

where negative sign indicates that work is done on the block

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(a)   v = 5.42m/s

(b)   vo = 4.64m/s

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(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.

You use the following formula:

h_{max}=\frac{v_o^2}{2g}       (2)

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v_o=\sqrt{2gh_{max}}=\sqrt{2(9.8m/s^2)(1.10m)}=4.64\frac{m}{s}

The velocity of the ball is 4.64m/s

(c) The acceleration is given by:

a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-(-5.42m)/s}{3.50*10^{-3}s}=2874.285\frac{m}{s^2}

a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-5.42m/s}{3.50*10^{-3}s}=-222.85\frac{m}{s^2}

The acceleration of the ball is 2874.28/s^2

(d) The compression of the ball is:

\Deta y=\frac{v^2}{2(a)}=\frac{(5.42m/s)^2}{2(2874.28m/s^2)}=5.11*10^{-3}m

THe compression of the ball when it strikes the floor is 5.11*10^-3m

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