When an object slides,there is less friction than when it rolls
1 answer:
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Answer:
Explanation:
Applied force, F = 18 N
Coefficient of static friction, μs = 0.4
Coefficient of kinetic friction, μs = 0.3
θ = 27°
Let N be the normal reaction of the wall acting on the block and m be the mass of block.
Resolve the components of force F.
As the block is in the horizontal equilibrium, so
F Cos 27° = N
N = 18 Cos 27° = 16.04 N
As the block does not slide so it means that the syatic friction force acting on the block balances the downwards forces acting on the block .
The force of static friction is μs x N = 0.4 x 16.04 = 6.42 N .... (1)
The vertically downward force acting on the block is mg - F Sin 27°
= mg - 18 Sin 27° = mg - 8.172 ... (2)
Now by equating the forces from equation (1) and (2), we get
mg - 8.172 = 6.42
mg = 14.592
m x 9.8 = 14.592
m = 1.49 kg
Thus, the mass of block is 1.5 kg.
The momentum increases by a factor of 2
Explanation:
We can solve this problem by rewriting the momentum of the rocket in terms of the kinetic energy and the mass.
The kinetic energy of the rocket is:
(1)
where
m is the mass
v is the velocity
The momentum of the rocket is
(2)
From eq.(1) we get
and substituting into (2),
Now in this problem we have:
- The kinetic energy of the rocket is increased by a factor 8:
- The mass is reduced by half:
Substituting, we find the new momentum:
So, the momentum increases by a factor of 2.
Learn more about momentum and kinetic energy:
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