February 22, 2008
PHY2053 Discussion
Quiz 5 (Chapter 5.3 -5.7)
Name:
UFID:
***1. (5 pts) A block starts from rest and slides 2.00 m down a 30.0º incline and then
slides some distance on a horizontal surface before coming to rest. The incline and the
horizontal surface have the same coefficient of kinetic friction against the block. Now
suppose we want to send the block back from its final position to its initial position.
What initial velocity we need to give the block?
First we apply the work-energy theorem to the motion of the block which slides down
the incline and comes to rest on the horizontal surface. We solve it for the horizontal
distance Δx:
-μmgcosθL-μmgΔx = -mgsinθL (L = 2m) ⇒ Δx = (Lsinθ-μLcosθ)/μ
We need to give the initial velocity to the block so that it stops at the original position.
Applying the work-energy theorem again, we have
-μmgcosθL-μmgΔx = mgsinθL-(1/2)mv²
Substitute the expression for Δx and solve for v:
-μmgcosθL-μmg(Lsinθ-μLcosθ)/μ = mgsinθL-(1/2)mv² ⇒ (1/2)mv² = 2mgsinθL
⇒ v = 2√(gsinθL) = 2√(9.8×sin30º ×2) = 6.26 m/s
*2. (5 pts) A 10.0 kg block is dropped from a height of 2.00 m above a light vertical
spring. The block lands on the spring and compresses it by a maximum of 0.500 m.
What is the spring constant?
The initial and final kinetic energy is both zero, and the total change in height is 2+0.5
= 2.5 m. Using the work-energy theorem, we have (h = 2 m, d = 0.5 m)
mg(h+d) = (1/2)kd² ⇒ k = 2mg(h+d)/d² = 2×10×9.8(2+0.5)/0.5² = 1.96×10³ N/m
**3. (5 pts) An 8.00 kg block situated on a 40.0º rough incline is attached to a string that
passes over a frictionless pulley at the top of the incline and then connected to a
horizontal spring. The other end of the spring is fixed and it has a spring constant of 200
N/m. The block is released from rest when the spring is unstretched. The block slides
0.300 m down the incline before coming to rest. Find the coefficient of kinetic friction
between block and incline.
When the block slides down Δx, its change in height is -Δxsinθ. Using the work-energy
theorem, we have
-μmgcosθΔx = -mgsinθΔx+(1/2)kΔx² ⇒ μ = tanθ-kΔx/(2mgcosθ)
= tan40º-200×0.3/(2×8×9.8cos40º) = 0.340
*4. (5 pts) An 800 kg elevator starts from rest, accelerates upward uniformly and
reaches its cruising speed of 2.50 m/s in 4.00 seconds. Find the average power of the
elevator motor during this period.
The average velocity and acceleration of the elevator are given by
V = (Vf+Vi)/2 = (2.5+0)/2 = 1.25 m/s & a = (Vf-Vi)/Δt = (2.5-0)/4 = 0.625 m/s².
We apply the Newton’s 2nd law to the elevator and solve it for the force the motor
provides on the elevator:
ma = F-mg ⇒ F = m(a+g) = 800×(0.625+9.8) = 8340 N
Thus, the power is given by
P = FV = 8340×1.25 = 1.04×10 w