Final answer:
a) The change in kinetic energy of the crate is 1189.89 J. b) The speed of the crate after being pulled the 8.33 m is approximately 14.34 m/s.
Step-by-step explanation:
a) To calculate the change in kinetic energy of the crate, we need to use the formula:
ΔKE = W + Q
where ΔKE is the change in kinetic energy, W is the work done by the 150 N force, and Q is the heat energy dissipated due to friction. Since the force is parallel to the incline, the work done can be calculated as:
W = Fd cos θ
Plugging in the values, we get:
W = (150 N)(8.33 m) cos 24° = 1189.89 J
Since the crate is pulled against friction, there is also heat energy dissipated. Let's assume that all of the work done by friction is dissipated as heat. Therefore, ΔKE = W = 1189.89 J.
b) To calculate the speed of the crate after being pulled the 8.33 m, we can use the equation:
vf² = vi² + 2ad
where vf is the final velocity, vi is the initial velocity (which is given as 1.5 m/s), a is the acceleration (which can be calculated as the net force divided by the mass of the crate), and d is the distance (which is 8.33 m). Plugging in the values, we get:
vf² = (1.5 m/s)² + 2((150 N) / (13 kg))(8.33 m) = 3.02 m²/s² + 202.56 m²/s² = 205.58 m²/s²
Taking the square root of both sides, we find the final speed:
vf ≈ √205.58 m²/s² ≈ 14.34 m/s