a 50 kg block is being pulled up a 37 degree incline with a force of 450 N applied parallel to the incline. The block moves at a constant velocity. What is the kinetic friction? power if the velocity is 1.5 m/s? acceleration if the applied force increases to 500N?

IF:-

F=450N, M=50kg, & V=1.5m/s
to find *Friction, *Power, & *Acceleration if the applied force increases to 500N
:-friction=applied force/normal force =f/450, f=mgsin€, =50*10sin€=300
:-friction=450/300=1.5
power=work/time, workdone=mg
:-50*10=500
acceleration if force is 500N
f=ma, a=f/m
:-500/50=10

To solve these problems, we'll need to break them down into smaller steps and use relevant equations. Let's go through each question one by one:

1. What is the kinetic friction?
First, we'll find the gravitational force acting on the block. The formula for gravitational force is given by:
F_gravity = m * g
where m is the mass (50 kg) and g is the acceleration due to gravity (typically taken as 9.8 m/s^2).

F_gravity = 50 kg * 9.8 m/s^2
F_gravity = 490 N

Next, we'll find the component of the gravitational force acting parallel to the incline. This can be calculated using the following formula:
F_parallel = F_gravity * sin(theta)
where theta is the angle of the incline (37 degrees).

F_parallel = 490 N * sin(37°)
F_parallel ≈ 294 N

Since the block is moving with a constant velocity, the applied force must equal the force of kinetic friction. So, the kinetic friction force can be found using the formula:
F_kinetic_friction = 450 N - F_parallel
F_kinetic_friction = 450 N - 294 N
F_kinetic_friction ≈ 156 N

Therefore, the kinetic friction force is approximately 156 N.

2. What is the power if the velocity is 1.5 m/s?
Power can be calculated using the formula:
Power = Force * Velocity

In this case, the force applied parallel to the incline is the same as the kinetic friction force, which is approximately 156 N. The given velocity is 1.5 m/s.

Power = 156 N * 1.5 m/s
Power = 234 Watts

Therefore, the power is 234 Watts.

3. What is the acceleration if the applied force increases to 500 N?
To find the acceleration, we can use Newton's second law of motion, given by the formula:
F_net = m * a

In this case, the net force (F_net) acting on the block is the sum of the applied force (500 N) and the component of the gravitational force acting parallel to the incline. We already calculated the latter as approximately 294 N.

F_net = 500 N - 294 N
F_net = 206 N

Using Newton's second law, we can now solve for the acceleration (a):
F_net = m * a

206 N = 50 kg * a

a = 206 N / 50 kg
a ≈ 4.12 m/s^2

Therefore, the acceleration is approximately 4.12 m/s^2 when the applied force increases to 500 N.