Answer :
1) Gravitational force: 98 N
2) Normal force: 98 N
3) Frictional force: 39.2 N
4) Acceleration: [tex]2.08 m/s^2[/tex]
Explanation:
1)
The gravitational force acting on an object is the force with which the object is pulled towards the Earth.
For an object near the Earth's surface, the gravitational force (also called weight of the object) is given by:
[tex]F_g=mg[/tex]
where
m is the mass of the object
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity at the Earth's surface
In this problem, we have a box of mass
m = 10 kg
Therefore, the weight of the box is:
[tex]F_g=(10)(9.8)=98 N[/tex]
2)
The normal force is the reaction force exerted by the floor on the box due to the push exerted by the box itself on the floor (these two forces represent the action-reaction pair of Newton's third law of motion).
The normal force is indicated with N.
The magnitude of this force can be found using Newton's second law along the vertical direction: in fact, we have two forces acting along this direction,
- The gravitational force, [tex]F_g[/tex], downward
- The normal force, N, upward
The box is in equilibrium along this direction, so the net force is zero; so we have:
[tex]N-F_g=0[/tex]
And so, the normal force is:
[tex]N=F_g=98 N[/tex]
3)
To frictional force is the resistive force acting on the box, opposite to the motion of the box, due to the roughness of the surfaces of the box and of the floor touching each other.
The magnitude of the frictional force for an object on a horizontal surface is given by:
[tex]F_f=\mu N[/tex]
where
[tex]\mu[/tex] is the coefficient of friction
N is the normal force
In this problem, we have
[tex]\mu=0.4[/tex]
N = 98 N (part b)
Therefore, the frictional force is
[tex]F_f=(0.4)(98)=39.2 N[/tex]
4)
The acceleration can be found applying Newton's second law of motion along the horizontal direction.
In fact, the net force along the horizontal direction must be equal to the product of the mass of the box and its acceleration. So we have:
[tex]F-F_f = ma[/tex]
where
F = 60 N is the horizontal push applied on the box
[tex]F_f=39.2 N[/tex] is the frictional force acting on the box, which resists the motion
m = 10 kg is the mass of the box
a is the acceleration of the box
Solving for a, we find the acceleration:
[tex]a=\frac{F-F_f}{m}=\frac{60-39.2}{10}=2.08 m/s^2[/tex]