InterviewSolution
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InterviewSolution
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A block of mass `m` is moving with a speed `v` on a horizontal rought surface and collides with a horizontal monted spring of spring constant `k` as shown in the figure .The coefficient of friction between the block and the floor is `mu` The maximum cobnpression of the spring is A. `- (mu mg)/(k) + (1)/(k) sqrt((mu mg)^(2) + mkv^(2))`B. ` (mu mg)/(k) + (1)/(k) sqrt((mu mg)^(2) + mkv^(2))`C. `- (mu mg)/(k) + (1)/(k) sqrt((mu mg)^(2) - mkv^(2))`D. `(mu mg)/(k) + (1)/(k) sqrt((mu mg)^(2) + mkv^(2))` |
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Answer» Correct Answer - A In pressent of friction both the spring force and the frictional act so as to oppose the compression of the spring Work done by the net force `W = -(1)/(2) kx_(m)^(2) - mu mhs_(m)` where `x_(m)` is the maximum compression of the spring change in kinetic energy `Delta K = K_(1) - K_(1) = 0 - (1)/(2) mv^(2)` Accerding to worik energy therem `W = Delta K` `- (1)/(2) kx_(m)^(2) - mu mgx_(m) = -(1)/(2)mv^(2)` ` kx_(m)^(2) + mu mgx_(m) = (1)/(2)mv^(2)` ` kx_(m)^(2) + 2mu mgx_(m) = -mv^(2)= 0` ` x_(m)^(2) +(2 mu mgx_(m))/(k) - (mv^(2))/(k) = 0` it is a quadratic equation in `x_(m)` Solving this equation for `x_(m)` is positive we get `x_(m) = -(mu mg)/(k) + (1)/(k) sqrt((mu mg)^(2) + mkv^(2))` |
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