Question

A stockroom worker pushes a box with mass 11.2 kg on a horizontal surface with a constant speed of 3.40 m/s . The coefIficient of kinetic friction between the box and the surface is 0.21.
a. What horizontal force must the worker apply to maintain the motion?
b. If the force calculated in part (a) is removed, how far does the box slide before coming to rest?

Answer 1

Answer:

a) 23.1 N b) 2.81 m

Explanation:

a)

• In order to keep the speed constant, the worker must apply an horizontal force, that be equal and opposite to the kinetic friction force.
• This is needed in order to the net force be zero, required condition for the box moves at a constant speed, according Newton's 2nd law:

        $F_{net} =m*a = 0$

        ⇒ a = 0 ⇒ v= constant

• The horizontal applied force must be as follows:

        $F = \mu_{k} * N$

• The normal force, in this particular case, is equal and opposite to the weight of the object, so it is directed directly upward:

       $N = m*g = 11.2 kg * 9.8 m/s2 = 110.0 N$

• Replacing by the values, we find the horizontal force needed as follows:

       $F = \mu_{k} * m*g\\\\ 0.21*11.2kg*9.8 m/s2= 23.1 N$

• The worker must apply an horizontal force of 23.1N (assumed the direction of the movement as the positive one) to maintain the motion.

b)

• If the force calculated in a) is removed, there will be a net force acting on the box.
• This net force, due to the kinetic friction force, will cause an acceleration (deceleration) that will slow down the box.
• We can find the value of this acceleration (assumed to be constant) just applying Newton's 2nd law to the box, subject to the friction force only:

       $F_{net} = m*a = 11.2 kg*a = -23.1N\\\\ a =\frac{-23.1N}{11.2kg} =-2.06m/s2$

• As we assumed that the acceleration is constant, in order to find the horizontal displacement prior to come to rest, we can use the following kinematic equation:

        $v_{f} ^{2} -v_{o} ^{2} = 2*a*d$

• In this equation, we have the following givens:
• vf = 0, v₀ = 3.4 m/s, a= -2.06 m/s²
• Replacing these values, we can solve for the displacement d as follows:

       $d =\frac{v_{o} ^{2} }{2*a} \\ \frac{(3.4m/s)^{2}}{2*2.06 m/s2} =2.81 m$

• If  the force calculated in a) is removed, before coming to rest, the box will slide through 2.81 m.