A body of mass 1kg is placed on a rough plank that is moving on a rough horizontal surface with a uniform velocity of 10m/s. find the frictional force on the body.(coefficient of static friction is 0.4 and coefficient of kinetic friction is 0.2)
Answers
Answer:
Explanation:
Here , maximum value of static friction that can act on the body = μ s mg
= 4 N
(i) Since the applied force is less than the maximum possible value of static friction
therefore , the force of friction will be just enough to stop the motion of the body.
Here , friction force = 15 N
(ii) Here, applied force = maximum possible value of static friction.
Friction force = 25 N
(iii) Here , applied force > maximum value of static friction.
Therefore , the body moves and kinetic friction acts on the body.
Friction force = μ
k
mg = 20 N.
Answer:
Fa−F0=ma
25−10=5a
a=3 m/s2
fms=210×5×10=10N
v2=2as=2×3×10=60
K.E=12mv2=12×5×60=150 J
(or)
Friction=μN
=0.2×5×10
Fnet=ma
25−10=5a⇒a=m/s2
v2=2as=2×3×10=60
KE=12mv2
KE=12×5×60
KE=150J