ANSWER:
F(h)= 230 N is the horizontal force you will need to move the pickup along the same road at the same speed.
STEP-BY-STEP EXPLANATION:
F(h) is Horizontal Force = 200 N
V is Speed = 2.4 m/s
The total weight increase by 42%
coefficient of rolling friction decrease by 19%
Since the velocity is constant so acceleration is zero; a=0
Now the horizontal force required to move the pickup is equal to the frictional force.
F(h) = F(f)
F(h) = mg* u
m is mass
g is gravitational acceleration = 9.8 m/s^2
200 = mg*u
Since weight increases by 42% and friction coefficient decreases by 19%
New weight = 1+0.42 = 1.42 = (1.42*m*g)
New friction coefficient = μ = 1 - 0.19 = 0.81 = 0.81 u
F(h) = (0.81μ) (1.42 m g)
= (0.81) (1.42) (μ m g)
= (0.81) (1.42) (200)
= 230 N
Answer:
Magnitude of the average force exerted on the wall by the ball is 800N
Explanation:
Given
Contact Time = t = 0.05 seconds
Mass (of ball) = 0.80kg
Initial Velocity = u = 25m/s
Final Velocity = 25m/s
Magnitude of the average force exerted on the wall by the ball is given by;
F = ma
Where m = 0.8kg
a = Average Acceleration
a = (u + v)/t
a = (25 + 25)/0.05
a = 50/0.05
a = 1000m/s²
Average Force = Mass * Average Acceleration
Average Force = 0.8kg * 1000m/s²
Average Force = 800kgm/s²
Average Force = 800N
Hence, the magnitude of the average force exerted on the wall by the ball is 800N
Answer:
10kg
Explanation:
Let PE=potential energy
PE=196J
g(gravitational force)=9.8m/s^2
h(change in height)=2m
m=?
PE=m*g*(change in h)
196=m*9.8*2
m=10kg