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Irina18 [472]
3 years ago
15

B)

Physics
1 answer:
Vlad1618 [11]3 years ago
5 0

Answer:

a) 16m/s b) 192m

Explanation:

v1=32m/s a=-2m/s^2 t=8s v2=? d=??

a) I will use this equation v2= v1 + a*t

v2= 32m/s + -2m/s^2 * 8s

v2= 32m/s + -16m/s

v2= 16m/s

b) v2^2=v1^2 + 2ad

rearranging

v2^2-v1^2=2ad

v2^2-v1^2/2= a d

v2^2-v1^2/2a=d

16m/s^2 - 32m/s^2/ 2 x-2m/s^2 =d

d=192m

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A single conservative force acts on a 4.50-kg particle within a system due to its interaction with the rest of the system. The e
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Answer: (a) The work done by this force on the particle is 42.71 J.

(b) The change in the potential energy of the system is -42.71 J.

(c) The kinetic energy the particle is 62.96 J.

Explanation:

(a)  For the given situation, expression for work done is as follows.

           W = \int_{0.9}^{5.15}Fdx

               = \int_{0.9}^{5.15}(2x + 4)dx

               = [2\frac{x^{2}}{2} + 4x]^{5.15}_{0.9}

               = (x^{2} + 4x)^{5.15}_{0.9}

               = [(5.15)^{2} + 4(5.15) - (0.9)^{2} - 4(0.9)]

               = 26.52 + 20.6 - 0.81 - 3.6

               = 42.71 J

Hence, the work done by this force on the particle is 42.71 J.

(b)  Expression for a conservative force is as follows.

              F = -\frac{dU}{dx}

            dU = -Fdx

      \int_{0.9}^{5.15}dU =  \int_{0.9}^{5.15}Fdx

        \int_{0.9}^{5.15}dU = -42.71 J

Therefore, the change in the potential energy of the system is -42.71 J.

(c) According to the work energy theorem,

         W = \Delta K.E

     K.E_{0.9} - K.E_{5.15} = W

            K.E_{0.9} = W + K.E_{5.15}

                          = 42.71 + \frac{1}{2}mu^{2}  

where, u = velocity of the mass at x = 0.9 m

         u = 3.0 m/s,       m = 4.50 kg

As,   K.E_{0.9} = W + K.E_{5.15}

                   = 42.71 + \frac{1}{2} \times 4.50 \times (3)^{2}

                   = 62.96 J

Therefore, the kinetic energy the particle is 62.96 J.

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