A block with mass 10 kg rests on a flat surface. The coefficient of static friction and kinetic friction between the two surfaces is 0.4 and 0.3 respectively. If the applied force is 50N find the follwing. a) what is the initial acceleration b) what is the acceleration of the block once it comes into motion. c) If the force is applied for 5s, what is the distance travelled in 5s? d) After the applied force is removed what is the distance travelled by the block before coming to stop. please help me.
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Answer:
a) 1 m/s^2
b) 2 m/s^2
c) 25m
d) 0 m
Explanation:
Net force = mass*acceleration
a) 50 - (0.4*100) = 10*a [Static frictional force comes into play]
a = 1 m/s^2
b) 50 - (0.3*100) = 10*a [Kinetic frictional force comes into play]
a = 2 m/s^2
c) u = 0 (Initial Velocity)
t = 5 s, a = 2 m/s^2
S = (u*t) + (1/2)*(a*t^2)
S = 0 + (1/2)*(2*(5^2))
S = 25 m
d) When the force is removed, Inertia force acting on block = m*a
= 10*2 = 20N
Which is less than kinetic frictional force, so the body comes to rest immediately.
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