2. A 14 kg box is sliding across a horizontal floor with an initial speed of 6 m
ID: 1584332 • Letter: 2
Question
2. A 14 kg box is sliding across a horizontal floor with an initial speed of 6 m/s. If the coefficient of friction between the box and the floor is 0.26, how far does the box slide before coming to rest? a) 35.3 f 3.2 2 b) 23.1 g) 13.4 c) 7.1 d) 32.2 e) 6.0 h) 21.2 i)21.2 j)10.1 3. In Fig. B, a block slides along a track that is frictionless except in the two places on the plateaus. At the bottom it slides a distance d in the friction until it stops. Find d (m). a) 5.54 f 0.50 b) 2.85 g) 0.67 c) 1.05 1.82 d) 0.32 i)2.03 e) 0.14 j)3.71Explanation / Answer
Q2:
Here kinetic energy change due to friction.
Change in kinetic energy (Delta KE) = workdone by the friction, Wf
friction force Ff = coefficient of friction (u) * normal force (N)
Normal force, N = mg
Therefore,
Ff = - u * mg
Wf = Ff * d
KEf - KEi = Wf
0.5*m*vf2 - 0.5*m*vi2 = Ff * d = - u * mg * d
0.5*vf2 - 0.5*vi2 = - u * g * d ----------------------->[1]
vf = 0; vi = 6 m/s; u = 0.26; g = 9.81 m/s2
from eqn[1],
d = 7.1 m [correct option is (c)]
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