Question

a 5 kg object is moving horizontally at 6 metre per second in order to change its speed 10 metre per second net work done on an object must be​

Answer 1

Answer:-

W = 160 J

Given :-

m = 5 kg

u = 6 m/s

v = 10 m/s

To find :-

The net work done on the object be.

Solution:-

• The net work done on the object is = Change in its kinetic energy.

• Kinetic energy of the body.

m = 5kg

u = 6 m/s

→$\mathsf{ K_e = \dfrac{mu^2}{2}}$

→$\mathsf{ K_e = \dfrac{5 \times (6)^2}{2}}$

→$\mathsf{K_e = \dfrac{5 \times 36}{2}}$

→$\mathsf{K_e = \dfrac{180}{2}}$

→$\mathsf{ K_e = 90 J}$

• Kinetic energy of the body when,

m = 5 kg

v = 10 m/s

→$\mathsf{K_e' = \dfrac{mv^2}{2}}$

→$\mathsf{K_e' = \dfrac{5 \times (10)^2}{2}}$

→$\mathsf{ K_e' = \dfrac{5 \times 100}{2}}$

→$\mathsf{K_e' = \dfrac{500}{2}}$

→$\mathsf{K_e' = 250 J }$

Now,

work done on the object is :-

→$\mathsf{ W = K_e' - K_e}$

→$\mathsf{W = 250 - 90}$

→$\mathsf{W = 160 J }$

hence,

The work done on the object is 160J.

Answer 2

Explanation:

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