Question

A 60-kilogram sled is coasting with a constant velocity of 10 m/s over smooth ice. It enters a rough stretch of ice 6.0 m longs in which the force of friction is 120 N. a) What is the acceleration during this stretch? B) With what speed dies it emerge from the rough patch?

Answer 1

The sled accelerates at 2 m/s² due to a frictional force of 120 N. To find the final velocity after the rough patch, use the work-energy principle rather than the equation of motion, as the time is not given.

Step-by-step explanation:

The question pertains to the frictional force acting on a sled as it moves over a patch of rough ice and the resulting effects on the sled's motion according to Newton's second law of motion.

Part a: Acceleration due to friction:

To find the acceleration of the sled during the rough patch, we use the formula a = F/m, where F is the force of friction and m is the mass of the sled. Considering the force of friction is 120 N and the sled's mass is 60 kg, the acceleration a is calculated as:

a = F/m = 120 N / 60 kg = 2 m/s²

Part b: Final velocity after rough patch:

To determine the sled's final velocity, we use the equation of motion v = u + at, where v is the final velocity, u is the initial velocity, and t is the time. As the sled is coasting at a constant velocity of 10 m/s before entering the rough patch and friction acts to decelerate it, v is less than the initial velocity u. Without the time t, we instead use the work-energy principle, considering the work done by the friction force over the distance of the rough patch, which is 6.0 m.

Answer 2

-2 m/s²

8.7 m/s

Step-by-step explanation:

Using Newton's second law, the sum of forces in the x direction:

∑F = ma

-120 N = (60 kg) a

a = -2 m/s²

Given:

Δx = 6.0 m

v₀ = 10 m/s

a = -2 m/s²

Find: v

v² = v₀² + 2aΔx

v² = (10 m/s)² + 2 (-2 m/s²) (6.0 m)

v = 8.7 m/s