16.7k views
2 votes
A block is given an initial speed of 3m/s up the 22.0 degree plane. What is the question?

User Itzhaki
by
8.2k points

1 Answer

2 votes

Final answer:

The block's acceleration down the ramp is 4.9 m/s^2 and the force of the ramp on the block is 9.8 N. The force applied upward along the ramp to allow constant velocity is also 9.8 N.

Step-by-step explanation:

The block's acceleration down the ramp can be determined using the formula:

Acceleration = gravitational acceleration * sin(angle) = 9.8 m/s^2 * sin(30°) = 4.9 m/s^2.

The force of the ramp on the block can be determined using the formula:

Force = mass * acceleration = 2.0 kg * 4.9 m/s^2 = 9.8 N.

To find the force applied upward along and parallel to the ramp that would allow the block to move with constant velocity, we can use the equation:

Force applied upward = force of gravity * sin(angle) = 2.0 kg * 9.8 m/s^2 * sin(30°) = 9.8 N.

User Ivan Baev
by
9.2k points