Neglect the flattening of the earth at the poles. 13) A block of mass is moving

Question

Neglect the flattening of the earth at the poles.
13) A block of mass is moving along the horizontalfrictionless surface with a speed of 5.70 m/s.    If the slope is11.00 and the coefficient of kinetic frictionbetween the block and the incline is 0.260,    how far does the block travel up theincline? (Figure 7.5)
14) Two masses, each 24 kg, are at rest and connected asshown. The coefficient of kinetic friction    between the inclined surface and the massis 0.31. Find the speed of the masses after they have    moved 1.6 m. (Figure 7.6) 15) A 475 gram ball is traveling horizontally at 12.0m/s to the left when it is suddenly struck       horizontally by a bat, causing it toreverse direction and initially travel at 8.50 m/s to theright.    If the bat produced an average forceof 1275 N on the ball, for how long (in ms) was it in contact    with the ball?

Explanation / Answer

Let the block travel distance s along the incline. Frictional force is kinetic and thus a constant Ffr =N = Mgcos11 = 0.260*m*9.8*cos11 N    Now, work done by friction = Wfr = Ffr . s =-Ffr*s    Also, work done by gravity = Fg . h = -Fg *s*sin11 = -m*9.8*s*sin11 This work done results in stoppage of block afterdistance s. So, Wnet = -0.260*m*9.8*s*cos11 N +(-m*9.8*s*sin11) = KE = 0 - 0.5*m*v*v => s = 0.5*5.7*5.7 / [ 0.260*9.8*cos11 +9.8*sin11]   m = .....................

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