The force pressing two stacked objects together is equal to the weight of the object on top. The friction between a piece of woo
1 answer:
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Answer:
a) a = 6.1 m/s^2
b) a = 0.98m/s^2
Explanation:
Mass of slab = 40kg
Mass of block = 10kg
Coefficient of static friction (Us) = 0.60
Kinetic coefficient (UK) = 0.40
Horizontal force = 100N
The normal reaction from 40kg slab on 10 kg block = 10*9.81
= 98.1N
Static frictional force = Us*R
= 98.1*0.6
= 58.86N
This is less than the force applied
If 10 kg block will slide on the 40 kg slab, net force = 100 - kinetic force
Kinetic force (Uk*R) = 0.4*98.1
= 39.28N
= 39N
Net force = 100 -39
= 61N
Recall that F = ma
For 10 kg block
a = F/m
a = 61/10
a = 6.1m/s^2
b) Frictional force on 40 kg slab by 10 kg = 98.1*0.4
= 39.24
= 39N
F = ma
a = F/m
For 40kg slab
a = 39/40
a = 0.98m/s^2
Answer:
There are finally 4 rotations per second.
Explanation:
If a trapeze artist rotates once each second while sailing through the air, and contracts to reduce her rotational inertia to one fourth of what it was. We need to find the final angular velocity. It is a case of conservation of angular momentum such that :
Let ,
and
So,
So, there are finally 4 rotations per second. Hence, this is the required solution.