MATHPHYS please help thank you in advance
1 answer:
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
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