Assertion :A work done by friction is always negative. Reason: If frictional force acts on a body its kinetic energy may decrease.

Statement -1 is false, Statement-2 is true

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Statement -1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1

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Statement -1 is true,Statement-2 is true; Statement -2 is not a correct explanation for Statement-1

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Statement -1 is true, Staternent-2 is false

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Solution

The correct option is A Statement -1 is false, Statement-2 is true
Work done by friction maybe positive, negative or zero.
When frictional force is opposite to velocity, kinetic energy will decrease.

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Similar questions

- 1 :

0 < x < y \Rightarrow {log}_{a} x > {log}_{a} y

a > 1

- 2 :

a > 1, {log}_{a} x

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} = (n + 2) 2^{n - 1} .

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} x^{r} = (1 + x)^{n} + n x (1 + x)^{n - 1} .

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} = (n + 2) 2^{n - 1} .

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} x^{r} = (1 + x)^{n} + n x (1 + x)^{n - 1} .

S_{1} = \sum_{j = 1}^{10} j (j - 1)^{10} C_{j}, S_{2} = \sum_{j = 1}^{10} j .^{10} C_{j}, S_{3} = \sum_{j = 1}^{10} j^{2} .^{10} C_{j} .

S_{3} = 55 \times 2^{9} .

S_{1} = 90 \times 2^{5} a n d S_{2} = 10 \times 2^{8}

S_{1} = \sum_{j = 1}^{10} j (j - 1)^{10} C_{j}, S_{2} = \sum_{j = 1}^{10} j^{10} C_{j} and S_{3} = \sum_{j = 1}^{10} j^{2}^{10} C_{j} .

S_{3} = 55 \times 2^{9}

S_{1} = 90 \times 2^{8} and S_{2} = 10 \times 2^{8}

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