Question

# An object of mass $$10$$ $$kg$$, moving with a velocity of $${ 5.6\ ms }^{ -1 }$$ on a surface, comes to rest after traveling a distance of $$16$$ $$m$$. Thena. coefficient of friction is $$0.1$$b. work-done against the friction is $$156.8$$ $$J$$c. retardation of the body is $$8.4$$ $${ ms }^{ -2 }$$.d. the block comes to rest in $$1$$ $$sec$$

A
a & b are correct
B
b & c are correct
C
c & d are correct
D
a, b & c are correct

Solution

## The correct option is A a & b are correctWe have, $$s=16 \ m, v=0$$$$v^{2}-u^{2}=2\mu gs$$$$\therefore u = \sqrt {2 \mu gs}$$$$\therefore \mu = \displaystyle \frac{5.6^2}{2\times 10\times 16}$$$$\therefore \mu = 0.098 \simeq 0.1$$Now,Work done,$$W= \mu mgs = 0.098 \times 10 \times 10 \times 16 = 156.8\ J$$Also,$$ma = \mu mg \Rightarrow a = 0.98 m/s^2$$$$v = u + at \Rightarrow 0 = 5.6 - 0.98 \times t \Rightarrow t = 5.71s$$Physics

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