Calculating Change in Kinetic Energy for a Crate Pulled Up an Incline

[BallerRegis]

A crate is pulled by a force (parallel to the
incline) up a rough incline. The crate has an
initial speed shown in the figure below. The
crate is pulled a distance of 9.17 m on the
incline by a 150 N force.
The acceleration of gravity is 9.8 m/s2 .
a) What is the change in kinetic energy of
the crate?
Answer in units of J

b) What is the speed of the crate after it is
pulled the 9.17 m?
Answer in units of m/s

For part A i tried
F-mg[sin theta - (friction x cos theta)] and got 135.232 but got wrong answer then i tried multiplying it by the distance and got it wrong.
Also i tried using F - [ μ m g cos θ ] and got it wrong
Someone please help me

 

[Delphi51]

The question appears to be incomplete. Do you know the angle, initial speed and coefficient of friction? If so, your calculation of
"F-mg[sin theta - (friction x cos theta)]"
looks like a good start. Here F = 150, friction = μmg. The result of the calc is the total force on the mass, which you can use in F = ma to find the acceleration. Once you have that, you can use accelerated motion formulas to find the Vf and then the Ek to complete part (a).

 

[BallerRegis]

ok i got part a which was

533.0427

how do i get part B
i tried

V= square root 2(553.0427)/11
but got the wrong answer

 

[Delphi51]

Didn't you find the Vf in part (a) ? Once you found the acceleration, you would have used d = Vi*t + .5*a*t² to find the time and then
Vf = Vi + a*t to get Vf. Only after knowing Vf are you in a position to do the energy calc for the (a) answer.

 

[asteriatic]

I would start by reviewing the given information and making sure that all the units are consistent. The force is given in Newtons, the distance in meters, and the acceleration of gravity in meters per second squared. This is important to ensure that the final answer is in the correct units of Joules (J) for energy.

Next, I would use the formula for kinetic energy, which is KE = 1/2 * m * v^2, where m is the mass of the crate and v is its velocity. Since the initial speed is given in the figure, I would use that as the initial velocity in the equation.

For part A, I would calculate the final kinetic energy by plugging in the mass of the crate (which is not given but can be assumed to be constant) and the final velocity, which can be calculated using the formula v^2 = u^2 + 2as, where u is the initial velocity, a is the acceleration (which can be calculated using F=ma), and s is the distance traveled.

For part B, I would use the same formula to calculate the final velocity, using the initial velocity, acceleration, and distance given in the problem.

It is also important to note that the frictional force and the angle of the incline may affect the calculations, so it is important to make sure that these factors are taken into account when solving the problem. I would also double-check my calculations and make sure to round to the correct number of significant figures.