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
Answer:
=3.5 m/s
Explanation:
y = x tanθ - 1/2 g x² / (u²cos²θ )
y = 0.25 , x = 0.5, θ = 40°
.25 = .50 tan40 - .5 x 9.8x x²/ u²cos²40
.25 = .42 - 2.0875/u²
u = 3.5 m / s.
The potential energy of a 30N ball on the ground will be zero. With respect to height, h. Potential energy will be calculated like this. P=mgh. So if its on the ground relatively speaking the h=0. Thus inputting into the above formula. P=0.
Answer:
The force would be the same in both cases - option C.
Explanation:
The change in momentum is known as an impulse. In the two cases under consideration, the change in momentum is the same, thus impulse for both cases is the same.
Impulse is the average force multiplied by time interval.
I = F(average)*ΔT. Where F(average) is the average force and ΔT is the time interval.
The average force in both cases is the same since the collision time is the same.
Thus option C is the correct answer.
