Uncategorized: Practice Problem & Solution
Starting from rest, a body slides down a (45^circ) inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is: (1988)
Solution Explained:
To solve this problem, we apply the core principles of Uncategorized. Understanding the underlying formula is key to arriving at the correct answer below:
Acceleration without friction is (a_1 = gsintheta). With friction, (a_2 = g(sintheta - mucostheta)). Since (s = frac{1}{2}at^2), (a_1 t_1^2 = a_2 t_2^2). Given (t_2 = 2t_1), so (a_1 t_1^2 = a_2 (2t_1)^2 = 4 a_2 t_1^2). Thus, (a_1 = 4a_2). (gsintheta = 4 g(sintheta - mucostheta)). (sintheta = 4sintheta - 4mucostheta). So, (4mucostheta = 3sintheta), giving (mu = frac{3}{4}tantheta). For (theta=45^circ), (mu = frac{3}{4} times 1 = 0.75).
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