Answer:
Explanation:
Force of friction
F = μ mg
μ is coefficient of friction , m is mass and g is acceleration due to gravity .
If f be the force applied to pull the sled , the horizontal component of force should be equal to frictional force
The vertical component of applied force will reduce the normal force or reaction force from the ground
Reaction force R = mg - f sin28.3
frictional force = μ R where μ is coefficient of friction
frictional force = μ x (mg - f sin28.3 )
This force should be equal to horizontal component of f
μ x (mg - f sin28.3 ) = f cos 28.3
μ x mg = f μsin28.3 + f cos 28.3
f = μ x mg / (μsin28.3 + cos 28.3 )
a )
work done by pulling force = force x displacement
f cos28.3 x d
μ x mg d cos28.3 / (μsin28.3 + cos 28.3 )
b ) Putting the given values
= .155 x 53.1 x 9.8 x 25.3 cos28.3 / ( .155 x sin28.3 + cos 28.3 )
= 1796.76 / (.073 + .88 )
= 1885.37 J
c )
Work done by frictional force
= frictional force x displacement
= - μ x (mg - f sin28.3 ) x d
= - μ x mgd + f μsin28.3 x d
= - μ x mgd + μsin28.3 x d x μ x mg / (μsin28.3 + cos 28.3 )
d )
Putting the values in the equation above
- .155 x 53.1 x 9.8 x 25.3 +
.155 x .474 x 25.3 x .155 x 53.1 x 9.8 /( .155 x .474 + .88)
= -2040.67 + 149.92 / .95347
= -2040.67 + 157.23
= -1883.44 J .
= 375 m/s