A crate slides down a ramp inclined at 26.0° from the horizontal. The normal force of the ramp on the crate is 710 N and the coefficient of kinetic friction between the crate’s bottom and the ramp’s surface is 0.330. Find the magnitude of the force of kinetic friction that the ramp exerts on the crate.
ahan, jimcao, chris, christopher, or whoever --
Your umpteen posts have been removed.
Once you write up YOUR THOUGHTS, please re-post, and someone here will be happy to comment.
To find the magnitude of the force of kinetic friction that the ramp exerts on the crate, we first need to find the weight of the crate.
The weight of an object can be calculated using the formula:
Weight = mass * gravity
Next, let's find the weight of the crate by using the normal force of the ramp on the crate. Since the normal force is equal to the force of gravity, we can set:
Weight = Normal force
Weight = 710 N
Now, we need to find the gravitational acceleration acting on the crate. This value is usually taken as 9.8 m/s².
Weight = mass * 9.8 m/s²
Solving for mass, we have:
mass = Weight / 9.8 m/s²
= 710 N / 9.8 m/s²
Next, we can calculate the force of kinetic friction using the formula:
Force of kinetic friction = coefficient of kinetic friction * Normal force
Plug in the given values:
Force of kinetic friction = 0.330 * 710 N
Now, calculate the force of kinetic friction:
Force of kinetic friction = 0.330 * 710 N
Therefore, the magnitude of the force of kinetic friction that the ramp exerts on the crate is approximately 234.3 N.