February 15, 2008
PHY2053 Discussion
Quiz 4 (Chapter 4.1-4.5 & 5.1-5.2)
Name:
UFID:
**1. (5 pts) A 4.00 kg block lies on a 45º incline and is attached to a light string that
passes over a pulley and is then fastened to an 8.00 kg hanging object. The coefficient of
kinetic friction between the incline and the block is 0.25. Find the acceleration of the
block.
Taking the positive x-axes in the directions of motion, the equations of motion for the
block and the hanging object are
ma = T-mgsinθ-μmgcosθ
Ma = Mg-T.
To remove T, we add these two equations and solve for the acceleration:
(m+M)a = Mg- mgsinθ-μmgcosθ ⇒ a = (Mg- mgsinθ-μmgcosθ)/(m+M)
= 9.8×(8-4sin45º-0.25×4cos45º)/12 = 3.65 m/s²
**2. (5 pts) A 2.00 kg block lies on the left edge of a plank which has a length of 3.00 m
and a mass of 10.0 kg. The coefficient of kinetic friction between the block and the plank
is 0.300 and the plank is on a frictionless horizontal floor. A constant horizontal force of
8.00 N is applied to the block and set it in motion. How long does it take for the block to
reach the right edge of the plank?
Applying Newton’s 2nd law to the horizontal motions of the block and plank, we get
ma = F – μmg ⇒ a = (F – μmg)/m = (8-0.3×2×9.8)/2 = 1.06 m/s²
MA = μmg ⇒ A =μmg/M = 0.3×2×9.8/10 = 0.588 m/s²
When the block reaches the right edge of the plank, the displacement of the block is
larger than that of the plank by L = 3.00 m. Thus
L = Δx-ΔX = (1/2)at² -(1/2)At² ⇒ t = √(2L/(a-A)) = √(2×3/(1.06-0.588)) = 3.57 s
**3. (5 pts) A 20.0 kg sledge is pulled at constant speed by a rope inclined at 20.0º above
the horizontal. The sledge moves a distance of 30.0 m on a horizontal surface. The
coefficient of kinetic friction between the sledge and the surface is 0.300. How much
work is done by the rope on the sledge?
Since the sledge moves at constant speed, its acceleration is zero. Thus we have
Y: 0 = n+Tsinθ-mg ⇒ n = mg-Tsinθ
X: 0 = Tcosθ-μn = Tcosθ-μ(mg-Tsinθ) ⇒ T = μmg/(cosθ+μsinθ)
= 0.3×20×9.8/(cos20º+0.3sin20º) = 56.4 N
Using the definition of work, we get
W = TcosθΔx = 56.4×cos20º×30 = 1.59×10³ J.
*4. (5 pts) A block of mass 5.00 kg slides 2.00 m down a frictionless 30.0º incline. Find
the change in kinetic energy of the block.
Only the gravity does work on the block. Using work-energy theorem, we have
ΔKE = Wg = mgsinθΔx = 5×9.8×sin30º×2 = 49.0 J