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:
[tex]F_{net} =m*a = 0[/tex]
⇒ a = 0 ⇒ v= constant
- The horizontal applied force must be as follows:
[tex]F = \mu_{k} * N[/tex]
- The normal force, in this particular case, is equal and opposite to the weight of the object, so it is directed directly upward:
[tex]N = m*g = 11.2 kg * 9.8 m/s2 = 110.0 N[/tex]
- Replacing by the values, we find the horizontal force needed as follows:
[tex]F = \mu_{k} * m*g\\\\ 0.21*11.2kg*9.8 m/s2= 23.1 N\\[/tex]
- 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:
[tex]F_{net} = m*a = 11.2 kg*a = -23.1N\\\\ a =\frac{-23.1N}{11.2kg} =-2.06m/s2[/tex]
- 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:
[tex]v_{f} ^{2} -v_{o} ^{2} = 2*a*d[/tex]
- 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:
[tex]d =\frac{v_{o} ^{2} }{2*a} \\ \frac{(3.4m/s)^{2}}{2*2.06 m/s2} =2.81 m[/tex]
- If the force calculated in a) is removed, before coming to rest, the box will slide through 2.81 m.
