A mass m = 14 kg is pulled along a horizontal floor with NO friction for a dista
ID: 1452911 • Letter: A
Question
A mass m = 14 kg is pulled along a horizontal floor with NO friction for a distance d =8.7 m. Then the mass is pulled up an incline that makes an angle = 27° with the horizontal and has a coefficient of kinetic friction k = 0.41. The entire time the massless rope used to pull the block is pulled parallel to the incline at an angle of = 27° (thus on the incline it is parallel to the surface) and has a tension T =62 N.
What is the speed of the block right before it begins to travel up the incline? ........m/s
What is the work done by friction after the block has traveled a distance x = 4.5 m up the incline? (Where x is measured along the incline.) .......J
What is the work done by gravity after the block has traveled a distance x = 4.5 m up the incline? (Where x is measured along the incline.).......J
How far up the incline does the block travel before coming to rest? (Measured along the incline.)......m
On the incline the net work done on the block is:
-Positive or Negative or Zero?
Explanation / Answer
tension T = m*a
accelaration a = T/m = 62/14 = 4.42 m/s^2
a = (v^2-u^2)/(2*S)
V = sqrt(2*a*S) = sqrt(2*4.42*8.7) = 8.76 m/s
---------------------------------------------
frictional force is fk = mu_k*m*g*cos(27) = 0.41*14*9.8*cos(27) = 50.12 N
work done by the friction is W1 = -50.12*4.5 = -225.54 J
Work done by the gravity is W2 = -m*g*sin(27)*4.5 = -14*9.8*sin(27)*4.5 = -280.3 J
net force is Fnet = T - 50.12-62.28 = 62-112.4 = -50.4 N
Fnet = m*a = -50.4
accelaration a = -50.4/14 = -3.6 m/s^2
a = (v^2-u^2)/(2*S)
-3.6 = (0^2-8.76^2)/(2*S)
S = 10.658 m
The net work done on the block is Negative
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