A4.00-kg box sits on a ramp that is inclined at 30°above the horizontal.The coefficient of kinetic friction between the box and
1 answer:
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Answer:
<em>The force that would be applied on the rope just to start moving the wagon is 122 N</em>
Explanation:
Frictional force opposes motion between two surfaces in contact. It is the force that must be applied before a body starts to move. Static friction opposes the motion of two bodies that are in contact but are not moving. The magnitude of static friction to overcome for the body to move can be calculated using equation 1.
F = μ x mg .............................. 1
where F is the frictional force;
μ is the coefficient of friction ( μs, in this case, static friction);
m is mass of the object and;
g is the acceleration due to gravity( a constant equal to 9.81 m/)
from the equation we are provide with;
μs = 0.25
m = 50 kg
g = 9.81 m/
F =?
Using equation 1
F = 0.25 x 50 kg x 9.81 m/
F = 122.63 N
<em>Therefore a force of 122 N must be applied to the rope just to start the wagon.</em>
Answer:
The velocity of the other fragment immediately following the explosion is v .
Explanation:
Given :
Mass of original shell , m .
Velocity of shell , + v .
Now , the particle explodes into two half parts , i.e .
Since , no eternal force is applied in the particle .
Therefore , its momentum will be conserved .
So , Final momentum = Initial momentum
The velocity of the other fragment immediately following the explosion is v .