Q: A block A of mass 0.5 kg can slide on a frictionless incline of angle 30o and length 0.8 m kept inside an elevator going up with uniform velocity 2m/s. Find the time taken by the block to slide down the length of the incline if it is released from the top of the incline. Take g = 10 ms-2.
If the elevator is not accelerating it is not part of the problem.
I bet you can do it ignoring the elevator.
By the way the mass of the block is also irrelevant because it cancels (remember feathers and rocks fall together in a vacuum)
To find the time taken by the block to slide down the length of the incline, we can use the principles of kinematics.
First, let's find the acceleration of the block as it slides down the incline. The force acting on the block is its weight, which can be decomposed into two components parallel and perpendicular to the incline.
The component of weight parallel to the incline is given by W_parallel = m * g * sin(theta), where m is the mass of the block, g is the acceleration due to gravity, and theta is the angle of the incline.
Substituting the given values:
m = 0.5 kg
g = 10 m/s^2
theta = 30 degrees, but we need to convert it to radians, so theta = 30 * (pi/180) radians.
Now we can find the parallel component of weight:
W_parallel = m * g * sin(theta)
W_parallel = 0.5 kg * 10 m/s^2 * sin(30 * (pi/180))
W_parallel = 0.5 kg * 10 m/s^2 * 0.5
W_parallel = 2.5 N
Since the incline is frictionless, the net force acting on the block is equal to the parallel component of weight.
Next, we can use Newton's second law to find the acceleration of the block:
F_net = m * a
a = F_net / m
a = 2.5 N / 0.5 kg
a = 5 m/s^2
Now that we have the acceleration, we can use the equations of motion to find the time taken by the block to slide down the incline.
We'll use the equation:
s = ut + (1/2)at^2
where s is the distance traveled by the block (length of the incline), u is the initial velocity (0 m/s as the block is initially at rest at the top of the incline), a is the acceleration, and t is the time taken.
Using the given values:
s = 0.8 m
u = 0 m/s
a = 5 m/s^2
Rearranging the equation, we get:
s = (1/2)at^2
2s = at^2
t^2 = (2s)/a
t^2 = (2 * 0.8 m) / 5 m/s^2
t^2 = 0.32 m / m/s^2
t^2 = 0.32 s^2
Taking the square root of both sides, we find:
t ≈ 0.57 s
Therefore, the time taken by the block to slide down the length of the incline, when released from the top, is approximately 0.57 seconds.