Answer: the area of the pool.
Explanation:
The work done to push the box is equal to the product between the force and the distance through which the force is applied:
In our problem, the force is F=1.4 N and the distance covered is d=150 m, so the work done by pushing the box is
Answer: 0.61
Explanation:
This is calculation based on friction.
Since the box rests on a flat surface, the force that exists between them is known as frictional force.
Since the friction is dynamic (velocity is not zero)
The frictional force = kinetic energy gained by the body.
Ff = 1/2mv^2
coefficient of kinetic friction × normal reaction = 1/2mv^2
Since normal reaction is equal to the weight(force acting along the vertical component)
Normal reaction= mg = 50 × 10 = 500N. Therefore,
coefficient of kinetic friction × 500 = 1/2×50×3.5^2
coefficient of kinetic friction = 50×3.5^2/1000
coefficient of kinetic friction= 0.61
Answer:
Part 1
20 N
Part 2
0.4 m/s²
Part 3
4 m/s
Explanation:
The force which pulls the sled right = 50 N
The friction force exterted towards left by the snow = -30 N
The mass of the sled = 50 kg
Part 1
The sum of the forces on the sled, F = 50 N + (-30) N = 20 N
Part 2
The acceleration of the sled is given as follows;
F = m·a
Where;
m = The mass of the sled
a = The accelertion
a = F/m
∴ a = (20 N)/(50 kg) = 0.4 m/s²
The acceleration of the sled, a = 0.4 m/s²
Part 3
The initial velocity of the sled, u = 2 m/s
The kinematic equation of motion to determine the speed of the sled is v = u + a·t
The speed, <em>v</em>, of the sled after t = 5 seconds is therefore;
v = 2 m/s + 0.4 m/s² × 5 s = 4 m/s.
