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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 $$16$$ $$m$$. Then
a. 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
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B
b & c are correct
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C
c & d are correct
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D
a, b & c are correct
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Solution

The correct option is A a & b are correct
We 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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