Answer:
2.51 m/s
Explanation:
Parameters given:
Angle, A = 33°
Mass, m = 90kg
Inclined distance, D = 2m
Force, F = 600N
Initial speed, u = 2.3m/s
From the relationship between work and kinetic energy, we know that:
Work done = change in kinetic energy
W = 0.5m(v² - u²)
We also know that work done is tẹ product of force and distance, hence, net work done will be the sum of the total work done by the force from the students and gravity.
Hence,
W = F*D*cosA - w*D*sinA
w = m*9.8 = weight
=> W = 600*2*cos33 - 90*9.8*2*sin33
W = 45.7J
=> 45.7 = 0.5*m*(v² - u²)
45.7 = 0.5*90*(v² - 2.3²)
45.7 = 45(v² - 5.29)
=> v² - 5.29 = 1.016
v² = 6.306
v = 2.51 m/s
The final velocity is 2.51 m/s
5... 589. 40....even though the zeros don't really count
Answer:
1. :uniformly accelerated motion
2. .0m/s
