Answer:a) 34.5 N; b) 24.5 N; c) 10 N; d) 1J
Explanation: In order to solve this problem we have to used the second Newton law given by:
∑F= m*a
F-f=m*a where f is the friction force (uk*Normal), from this we have
F= m*a+f=5 Kg*2 m/s^2+0.5*5Kg*9.8 m/s^2= 34.5 N
then f=uk*N=0.5*5Kg*9.8 m/s^2= 24.5N
the net Force = (34.5-24.5)N= 10 N
Finally the work done by the net force is equal to kinetic energy change so
W=∫Force net*dr= 10 N* 0.1 m= 1J
The formula to calculate the coefficient of friction is μ = f÷N. The friction force, f, always acts in the opposite direction of the intended or actual motion, but only parallel to the surface.
HOPE THIS HELPS :)
This year is 60 years since I learned this stuff, and one of the things I always remembered is the formula for the distance a dropped object falls:
D = 1/2 A T²
Distance = (1/2) (acceleration) (time²)
The reason I never forgot it is because it's SO useful SO often. You really should memorize it. And don't bury it too deep in your toolbox ... you'll be needing it again very soon. (In fact, if you had learned it the first time you saw it, you could have solved this problem on your own today.)
The problem doesn't tell us what planet this is happening on, so let's make it easy and just assume it's on Earth. Then the 'acceleration' is Earth gravity, and that's 9.8 m/s² .
In 5 seconds:
D = 1/2 A T²
D = (1/2) (9.8 m/s²) (5 sec)²
D = (4.9 m/s²) (25 sec²)
D = 122.5 meters
In 6 seconds:
D = 1/2 A T²
D = (1/2) (9.8 m/s²) (6 sec)²
D = (4.9 m/s²) (36 sec²)
D = 176 meters