Question

MATHPHYS please help thank you in advance

Answer 1

Answer:

1. -6.43 J

2. 6.43 J

3. -12.25 J

4. 196 J

Explanation:

For the box on the ramp, there are two forces:

Weight force mg pulling down

Normal force N pushing perpendicular to the ramp

Sum of the forces in the parallel direction:

∑F = ma

-mg sin θ = ma

a = -g sin θ

a = -9.8 m/s² sin 15°

a = -2.54 m/s²

1. The distance the box moves is:

Δx = vt − ½ at²

Δx = (0 m/s) (1 s) − ½ (-2.54 m/s²) (1 s)²

Δx = 1.27 m

The work done by gravity is:

W = Fd

W = (-mg sin θ) (Δx)

W = -(2 kg) (9.8 m/s²) (sin 15°) (1.27 m)

W = -6.43 J

Notice the answer is negative.  That's because the force vector and displacement vectors are in opposite directions

2. The distance the box moves is:

Δx = v₀t + ½ at²

Δx = (0 m/s) (1 s) + ½ (-2.54 m/s²) (1 s)²

Δx = -1.27 m

The work done by gravity is:

W = Fd

W = (-mg sin θ) (Δx)

W = -(2 kg) (9.8 m/s²) (sin 15°) (-1.27 m)

W = 6.43 J

3. Final KE = initial KE + work by gravity

KE = KE₀ + W

0 = ½ mv² + W

W = -½ mv²

W = -½ (2 kg) (3.50 m/s)²

W = -12.25 J

4. Initial PE = final PE + work by gravity

PE₀ = PE + W

mgh = 0 + W

W = mgh

W = (2 kg) (9.8 m/s²) (10 m)

W = 196 J