A block of mass 1 kg, moving along x with speed = 10 m/s enters a rough region ranging from x = 0.1 m to x = 1.9 m. The retarding force acting on the block in this range is = – kx N, with k = 10 N/m. Then the final speed of the block as it crosses rough

Question

A block of mass 1 kg, moving along x with speed $v_{i}$ = 10 m/s enters a rough region ranging from x = 0.1 m to x = 1.9 m. The retarding force acting on the block in this range is $F_{r}$ = – kx N, with k = 10 N/m. Then the final speed of the block as it crosses rough region is [2025]

Smart Achievers · Accepted Answer

(4)

$v_{i}$

$\int_{10}^{v} v d v = - 10 \int_{0.1}^{1.9} x d x$

$\frac{v^{2} - 100}{2} = - 10 (\frac{{(1.9)}^{2} - {(0.1)}^{2}}{2})$

$\Rightarrow v = 8 m / s$

Solution

Q.
A block of mass 1 kg, moving along x with speed $v_{i}$ = 10 m/s enters a rough region ranging from x = 0.1 m to x = 1.9 m. The retarding force acting on the block in this range is $F_{r}$ = – kx N, with k = 10 N/m. Then the final speed of the block as it crosses rough region is [2025]

1 10 m/s

2 4 m/s

3 6 m/s

4 8 m/s

Ans.
(4)

$a = \frac{F}{m} = - 10 x \Rightarrow v \frac{d v}{d x} = - 10 x$

$\int_{10}^{v} v d v = - 10 \int_{0.1}^{1.9} x d x$

$\frac{v^{2} - 100}{2} = - 10 (\frac{{(1.9)}^{2} - {(0.1)}^{2}}{2})$

$\Rightarrow v = 8 m / s$

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