Solve Block on an Incline: Net Force, Accel, Speed in 9 m/s

[cstout]

[SOLVED] Block on an incline

A 11.4-kg block is released from rest on a frictionless track inclined at an angle of 47°. (a) What is the net force on the block after it is released? (b) What is the acceleration of the block? (c) If the block is released from rest, how long will it take for the block to attain a speed of 9 m/s?

I have been working on this problem but for some reason I must have an error somewhere, can anyone show me the correct answers.

 

[andrevdh]

If you show us your solution we can maybe find the error.

 

[Moetasim]

(a) The net force on the block is equal to its weight component down the incline, which can be calculated using the formula F = mgsinθ, where m is the mass of the block, g is the acceleration due to gravity (9.8 m/s^2), and θ is the angle of the incline. Plugging in the values, we get F = (11.4 kg)(9.8 m/s^2)sin(47°) = 79.3 N.

(b) The acceleration of the block can be calculated using Newton's second law, F = ma. Plugging in the net force calculated in part (a), we get 79.3 N = (11.4 kg)a. Solving for a, we get a = 6.96 m/s^2.

(c) To find the time it takes for the block to reach a speed of 9 m/s, we can use the formula v = u + at, where v is the final velocity, u is the initial velocity (which is 0 m/s in this case), a is the acceleration calculated in part (b), and t is the time. Plugging in the values, we get 9 m/s = 0 + (6.96 m/s^2)t. Solving for t, we get t = 1.29 seconds. Therefore, it will take the block 1.29 seconds to reach a speed of 9 m/s.