Explanation:
Given that,
The slope of the ramp, 
Mass of the box, m = 60 kg
(a) Distance covered by the truck up the slope, d = 300 m
Initially the truck moves with a constant velocity. We know that the net work done on the box is equal to 0 as per work energy theorem as :

u and v are the initial and the final velocity of the truck
(b) The work done on the box by the force of gravity is given by :

Here, 


W = -24550.13 J
(c) What is the work done on the box by the normal force is equal to 0 as the angle between the force and the displacement is 90 degrees.
(d) The work done by friction is given by :


Hence, this is the required solution.
Answer:
7.0 s, 69 m/s
Explanation:
If we take down to be positive, then the time to reach the ground is:
x = x₀ + v₀ t + ½ at²
240 m = (0 m) + (0 m/s) t + ½ (9.8 m/s²) t²
t = 7.0 seconds
The final velocity is:
v² = v₀² + 2a(x - x₀)
v² = (0 m/s)² + 2(9.8 m/s²) (240 m - 0 m)
v = 69 m/s
It would be spectral class M
hope this helps
we know the equation for the period of oscillation in SHM is as follows:
T = 2 * pi * sqrt(mass/k)
we know f = 1/T, so f = 1/(2 * pi) * sqrt(k/m).
since d = v*T, we can say v = d/t = d * f
the final equation, after combining everything, is as follows:
v = d/(2 * pi) * sqrt(k/m)
by plugging everything in
v = .75/(2 * pi) * sqrt((1 * 10^5)/(30))
We find our velocity to be:
v = 6.89 m/s
Answer:
v = 17.15 m/s
Explanation:
given,
angle of ramp = 17.0°
length of ramp(l) = 30 m
height of the ramp =


h = 15 m
using energy of conservation





v = 17.15 m/s
speed of block reaching at the bottom = v = 17.15 m/s