Use the work-energy theorem to find the velocity of the block when it's released by the spring. The work done by the spring on the block as it's restored to equilibrium is
W = 1/2 kx ²
where k is the spring constant and x is the compression of the spring. So
W = 1/2 (1400 N/m) (0.170 m)² = 20.23 J
This is equal to the block's change in kinetic energy ∆K,
W = ∆K
and since it starts from rest, the initial K is zero, leaving us with
W = 1/2 mv ²
where m is the mass of the block and v is its speed, so that
20.23 J = 1/2 (0.200 kg) v ²
==> v ≈ 14.2 m/s
The block slides at this speed across the frictionless surface until it hits the incline which introduces friction.
First, you need to find the length of the incline. It forms a 45° angle, and the underlying 45°-45°-90° triangle has a hypotenuse of length √2 (2.0 m) ≈ 2.83 m.
Next, you need to find the total work done on the block as it slides up the incline. Use Newton's second law to examine the forces acting on the block during this phase:
• the net force acting on the block in the direction perpendicular to the incline is
∑ F = n - mg cos(45°) = 0
where n = mg cos(45°) ≈ 1.39 N is the magnitude of the normal force and mg cos(45°) ≈ 1.39 N is the perpendicular component of the block's weight;
• the net force acting on the block parallel to the surface is
∑ F = -f - mg sin(45°) = ma
where f = µn = 0.210n ≈ 0.291 N is the magnitude of kinetic friction, mg sin(45°) ≈ 1.39 N is the parallel component of the weight, and a is the acceleration of the block.
Only the parallel forces do work on the block, and this work is negative because friction and weight oppose the block's sliding up the incline. The total work done on the block is then
W = (-0.291 N - 1.39 N) (2.83 m) ≈ -4.74 J
Use the work-energy theorem again to find the block's new speed v at the top of the incline:
W = ∆K
==> -4.74 J = 1/2 (0.200 kg) v ² - 1/2 (0.200 kg) (14.2 m/s)²
==> v ≈ 12.4 m/s
And now this becomes a projectile problem. The block travels a horizontal distance x after being launched at an angle of 45° with initial speed 12.4 m/s after time t according to
x = (12.4 m/s) cos(45°) t
Its height y from the 2.0 m-high surface at time t is given by
y = (12.4 m/s) sin(45°) t - 1/2 gt ²
The block lands on the surface when y = 0, which occurs after t ≈ 1.79 s, at which point the block has covered a distance d ≈ 15.7 m.
