A 0.350 kg block slides on a frictionless, horizontal surface with a speed of 1.
ID: 2160855 • Letter: A
Question
A 0.350 kg block slides on a frictionless, horizontal surface with a speed of 1.37 m/s. The block encounters an unstretched spring and compresses it 25 cm before coming to rest.(a) What is the force constant of this spring?
Your response differs from the correct answer by more than 10%. Double check your calculations. N/m
(b) For what length of time is the block in contact with the spring before it comes to rest?
Your response differs from the correct answer by more than 10%. Double check your calculations. s
(c) If the force constant of the spring is increased, does the time required to stop the block increase, decrease, or stay the same?
stays the same
decreases
increases
Explain.
Explanation / Answer
a) When the block comes to erst compressing the spring, the entire kinetic energy of the block gets converted into potential energy of the spring. (Using the Law of Conservation of Energy) 1/2 mv^2 = 1/2 kx^2 Substitute the values and solve for k. k = 10.51N/m b) Force exerted by the spring when it is compressed by a distance x = -kx Acceleration, a = - kx / m Now a = dv / dt = dv / dt * dx / dx = dv / dx * dx / dt = v dv / dx So v dv = - (kx/m) dx Integrate both sides (LHS within the limits u to v and RHS within the limits 0 to x; u is the initial velocity = 1.37 m/s) x^2 = m/k (u^2 - v^2) or x = sqrt(m/k) * sqrt(u^2 - v^2) .................(1) Also a = dv / dt So - kx / m = dv / dt Substitute the value of x from (1): dv / dt = - sqrt(k/m) * sqrt(u^2 - v^2) or dv = - sqrt(k/m) * sqrt(u^2 - v^2) * dt Integrate LHS within the limits u to v and RHS within the limits 0 to t : Simplify to get t = sqrt (m / k) [ sin^-1 (1) - sin^-1 (v / u) ] Put v (final velocity) = 0 t = sqrt (m / k) [ sin^-1 (1) - sin^-1 (0) ] Substitute sin^-1 (1) = pi/2 and sin^-1 (0) = 0 (Actually sin^-1 (x) can have multiple values but here you need to find the minimum value of the expression " sin^-1 (1) - sin^-1 (0) ") t = sqrt (m / k) * pi / 2 ........................(2) = 0.286 s c) From (2), we can see that t is inversely proportional to sqrt(k) So if we increase k, t will decrease.
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