Pete slides a crate up a ramp at an angle of 25◦ by exerting a 290 N force parallel to the ramp. The crate moves at a constant speed. The coefficient of friction is 0.28.
How much work did Pete do when the crate was raised a vertical distance of 2.24 m?
Answer in units of J
d = 2.24m/sin25 = 5.30 m Parallel to the
ramp.
Work = Fex * d = 290 * 5.30 = 153t J.
Well, Pete did quite a bit of work, but the real question is, did he do the work with a smile on his face? Let's find out!
To calculate the work done by Pete, we need to consider the force he exerted parallel to the ramp. This force can be broken down into two components: the force parallel to the ramp (290 N) and the force perpendicular to the ramp (290 N * sin(25°)). The force perpendicular to the ramp is what balances out the gravitational force acting on the crate.
Now, the work done by Pete can be calculated by multiplying the force parallel to the ramp by the distance the crate was raised vertically. So, the work done is:
Work = Force parallel to the ramp * Distance
= 290 N * 2.24 m
But wait, we're not done yet! We also need to take into account the coefficient of friction (0.28) and the work done against it. The force of friction can be calculated by multiplying the coefficient of friction by the normal force, which is equal to the force perpendicular to the ramp.
Frictional force = Coefficient of friction * Normal force
= 0.28 * (290 N * cos(25°))
Finally, we need to subtract the work done against friction from the total work done by Pete:
Total Work = Work - Frictional force * Distance
= (290 N * 2.24 m) - (0.28 * (290 N * cos(25°)) * 2.24 m)
So, the total work done by Pete when the crate was raised a vertical distance of 2.24 m is the final answer. I'll let you crunch the numbers and find out exactly how much work that is, my friend!
To calculate the work done by Pete in raising the crate, we need to consider the forces acting on the crate and the distance it is moved vertically.
First, let's find the force of friction acting on the crate. The force of friction can be determined using the equation:
Force of friction = coefficient of friction x normal force
The normal force is the force exerted by the crate perpendicular to the ramp, which is equal to the weight of the crate. The weight of the crate can be calculated using the formula:
Weight = mass x gravity
Since the crate moves at a constant speed, the force of friction will be equal in magnitude and opposite in direction to the force applied by Pete (290 N). Therefore, the force of friction is 290 N.
Next, we can calculate the work done by Pete using the formula:
Work = force x distance x cosine(theta)
Where:
force = 290 N (force applied by Pete)
distance = 2.24 m (vertical distance raised)
theta = 25 degrees (angle of the ramp)
Since the force and distance are in the same direction, the angle between them is 0 degrees, so the cosine(theta) will be equal to 1.
Work = 290 N x 2.24 m x cos(0)
Work = 290 N x 2.24 m x 1
Work = 648.8 J
Therefore, Pete did approximately 648.8 Joules (J) of work when the crate was raised a vertical distance of 2.24 m.
To find the work done by Pete in raising the crate, we need to calculate the net force acting on the crate, and then multiply it by the displacement of the crate.
First, let's find the force exerted by Pete parallel to the ramp. We are given that Pete exerts a force of 290 N parallel to the ramp. Since the ramp is inclined at an angle of 25 degrees, the component of this force parallel to the ramp can be calculated using trigonometry.
Force parallel to the ramp = Force exerted by Pete * cos(angle of the ramp)
Force parallel to the ramp = 290 N * cos(25 degrees)
Next, let's find the frictional force acting on the crate. The frictional force can be calculated using the equation:
Frictional force = coefficient of friction * Normal force
The normal force can be calculated using the weight of the crate, which is given by:
Weight of the crate = mass of the crate * acceleration due to gravity
Since the crate is moving at a constant speed, we know that the net force acting on it is zero. Thus, the force parallel to the ramp must be equal to the frictional force.
Now, let's find the displacement of the crate when it is raised vertically by 2.24 m.
Finally, we can calculate the work done by Pete using the equation:
Work done = Net force * displacement
Plug in the values and calculate to find the answer in joules.