4 kg steel ball accelerated to 62 m/s and experimental train kinematics

[jai6638]

A steel ball has a mass of 4 kg and rolls along a smooth level surface at 62 m/s.

a) Find its kinetic energy ( i found this.. )

b) At first, the ball was at rest on the surface. A constant force acted on it through a distance of 22 m to give it the speed of 62 m/s. What was the magnitude of the force?

so here's my thought process:

.5 mv^2 + mgh = .5mv^2 + mgh
0 = .5mv^2+mgh
-.5mv^2=mgh
mg=- .5mv^2/h
mg= - .5 x 4 x 62^2/22
mg= -350 N


is the above method correct?

Q2) In the 1950's, an experimental train that had a mass of 2.50x10^4 was powered across a level track by a jet engine that produced a thrust of 5x10^5 N for adistance of 509 m.

a) Find the work done on the train. ( found this to be 2.545E8)
b) Find the change in kinetic energy. ( 2.545E8)
c) Find the final kinetic energy of the train if it started from rest. ( how do i find this?? .5 mv^2 won't wrk since i don't know what the velocity is )
d) find the final speed of the train if there were no friction ( how do i find this?? )

thanks much

 

[Pyrrhus]

On your b) you should have used

$$\sum_{i=1}^{n} W_{i} = \Delta K$$

 

[jai6638]

On your b) you should have used

$$\sum_{i=1}^{n} W_{i} = \Delta K$$

so ur saying i would use W= delta K = fd?

but i'd still get the same answer of 350 N...

 

[Pyrrhus]

Yes, but i don't know where you got Potential Gravitational energy, the problem doesn't refers to a not leveled surface, plus there's inconsistency, because you got mg = -350N when that is weight, and the weight of the object is about 39.2 N.

 

[jai6638]

is the magnitude of force = 1.69 x 10^5 n??

thanks

 

[futb0l]

Q2) In the 1950's, an experimental train that had a mass of 2.50x10^4 was powered across a level track by a jet engine that produced a thrust of 5x10^5 N for adistance of 509 m.

a) Find the work done on the train. ( found this to be 2.545E8)
b) Find the change in kinetic energy. ( 2.545E8)
c) Find the final kinetic energy of the train if it started from rest. ( how do i find this?? .5 mv^2 won't wrk since i don't know waht the velocity is )
d) find the final speed of the train if there were no friction ( how do i find this?? )

thanks much

c) that's easy - it just means that KEi = 0 .. when KEi = 0, KEf = Work!
d) well then .5v^2 = work and then u can find v

 

[jai6638]

c) that's easy - it just means that KEi = 0 .. when KEi = 0, KEf = Work!
d) well then .5v^2 = work and then u can find v

so c= 2.545 x 10^8 right?