<span>By algebra, d = [(v_f^2) - (v_i^2)]/2a.
Thus, d = [(0^2)-(15^2)]/(2*-7)
d = [0-(225)]/(-14)
d = 225/14
d = 16.0714 m
With 2 significant figures in the problem, the car travels 16 meters during deceleration.</span>
Answer:
vB = 15.4 m/s
Explanation:
Principle of conservation of energy:
Because there is no friction the mechanical energy is conserve
ΔE = 0
ΔE : mechanical energy change (J)
K : Kinetic energy (J)
U: Potential energy (J)
K = (1/2)mv²
U = m*g*h
Where :
m: mass (kg)
v : speed (m/s)
h : hight (m)
Ef - Ei = 0
(K+U)final - (K+U)initial =0
(K+U)final = (K+U)initial
((1/2)mv²+m*g*h)final = ((1/2)mv²+m*g*h)initial , We divided by m both sides of the equation:
((1/2)vB² + g*hB = (1/2 )vA²+ g*hA
(1/2) (vB)² + (9.8)*(14.7) = 0 + (9.8)(26.8 )
(1/2) (vB)² = (9.8)(26.8 ) - (9.8)*(14.7)
(vB)² = (2)(9.8)(26.8 - 14.7)
(vB)² = 237.16

vB = 15.4 m/s : speed of the cart at B
Work = force * distance.
We must produce twice as much energy as we are lifting the weight twice as high.
But we are not increasing the force so we must increase the length of the ramp ( distance ) instead.
The new length will be twice as great as the previous length.
So 8 metres is required.
25 kg * 8 m = work = 100 kg * 2 m