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Answer:
1.40 N
Explanation:
The magnitude of the frictional force is given by:
where
is the coefficient of friction
N is the magnitude of the normal reaction
The coefficient of friction for this problem is . The magnitude of the normal reaction is equal to the combined weight of the boy and the sled, because the surface is horizontal, so
Therefore, the frictional force is
Answer:
A:- 50 J
B:- 500 J
Explanation:
a) Given that a 25 N force is applied to move the box. Also the floor is having friction surface.
So in order to move the box, the floor should have friction of atleast 25 N.
∴ friction = 25 N
Work done = force * displacement of box
Given, the box is moved 2 m across the floor
So, Work done = friction * 2 m
= 25 * 2
= 50 J
b) Given, the box is having weight of 250 N weight
Gravitational force is acting on the box which is equal to (mass * gravity)
∴ Force = 250 N
The box is lifted 2 m above the floor.
So, displacement = 2 m
Work done = force * displacement of box
Work done = 250 * 2
= 500 J
From the basic "heat lost by hot object=heat gained by colder object" principle, we have
m1c1ΔT1=m2c2ΔT2
where m1= 1kg
m2=4kg
c1=900J/kg k
c2=4200J/kg k
With this information at hand we have
m1c1(90-T)=m2c2(T-25)
after substituting the given values we can find that
T=28.3^{0}c