Calculating Kinetic Energy of Fridge Pushed 5m

[AHUGEMUSHROOM]

Q: A young Mr. Watford applies a force of 180N to a refrigerator on. kitchen floor.

Part (A):Find the net force on the refrigerator if the floor has a constant frictional force of 35N.

Ans:145N

Part (B): Calculate the acceleration of the refrigerator if it has a mass of 70kg.

Ans: 2.57 ms-2

Part (C) Calculate the kinetic energy of the fridge after it has been pushed 5.00m.

initial velocity = 0 ms-1
displacement = 5m
acceleration = 2.57 ms-2
mass = 70kg
time = ?s = ut + 1/2at^2
v = u + at
Ek=1/2 mv^2Substituting the given units into s = ut + 1/2at^2, I got time = 1.97s

Substituting the units into v = u + at, I got final velocity = 5.06 ms-1

And finally subbing the units into Ek= 1/2mv^2, I got Kinetic Energy = 896.13J.

I got one benefit of the doubt mark out of four possible marks, and if that was in an exam I would've got 0 marks as there are no benefit of the doubt marks.

Please reply asap I have an exam tomorrow.

Thanks

*ANSWERED*

E = W
W= Fs
=180 x 5 = 900J

 

[SteamKing]

Give us the entire problem statement, and you might get some constructive suggestions.

 

[AHUGEMUSHROOM]

Give us the entire problem statement, and you might get some constructive suggestions.

Fixed! :)

 

[gneill]

Rethink your calculation for the acceleration. The mass of the fridge will respond to the net force applied to it.

 

[rwisz]

KE = 1/2mv^2
= 1/2 x 70 x (5.06)^2 = 896.13J

The kinetic energy of the fridge after being pushed 5m is 896.13J.