Kinetic energy?

Question

A crate is pulled a force ( parallel to the incline) up a rough incline. The crate has an initial speed of 1.53 m/s The crate's mass is 11 kg. The crate is pulled a distance of 7.46 m on the incline by a 150 N force. The coefficient of kinetic friction is 0.347 between the crate and the incline. The acceleration of gravity is 9.8 m/s2.

a) What is the change in kinetic energy of the crate ?
b) What is the speed of the crate after it is pulled the 7.46 m ?

The mass of the crate is 11 kg and the angle is 21 degrees. Sorry for the missing details.

Answer 1

It's OK, you only forgot to put the angle, the other guy missed the 11 kg.

first, let's find F_frict,

F_normal=m*g*cos21=11*9.81*cos(21) =100.74 N
( force of gravity normall to incline )

and F_parallel=11*9.81*sin(21) = 38.67
( force of gravity parallel to incline )

F_frict=0.347*F_normal = 34.96 N

so F_incline= 150 - F_parallel - F_frict = 76.37 N

a)

Change of KE = Work = F_incline*d = 76.37*7.46 = 569.7202 J

b)

0.5*m*v^2 = 569.72, where V is change in speed
v=10.17769397 m/s

so the new speed is= 10.17769397 + 1.53 = 11.7 m/s

Answer 2

question is incomplete slope of the incline and mass of crate have to be given. Let mass be m and inclination of slope to horizontal, x

Change in kinetic energy = net force x distance traveled in the direction of net force or

KE = [150 - (9.8 x m x sine x) - (347 x 9.8 x m x cosine x)] x 7.46

speed = sqrt{2 KE/m}