
Help with physics question. See if I did right. Keep getting wrong answer.?
A student pulls a box of books with a force of 120N [E]. The box of has a mass of 20kg. The coefficient of kinetic friction between the box and floor is 0.5. a) Find the acceleration of box.
b) What if he pulls with a force of 120N [E30N].
a)
Fg = Force Gravity
Fn = Force normal
a = acceleration
Fnet = Net force
Ff = Frictional force
u = Coefficient
Fg = m x g
= 20 x 9.8
=192N
Fa = 120N
Ff = u x Fg
=0.5 x 192 = 96N
Fnet = Fa – Ff
Fnet = 120N – 96N
=24N[E]
Acceleration
F = m x a
a = F/m
a = 24/20 = 1.2m/s^2
But the answer is 1.1m/s^2
Someone please help me identify the problem.
b) I don’t understand what to do with this one. Can some please help me?
The first step with a problem like this is to draw a free body diagram, but since I don’t know how to draw in the given format I will describe the diagram to the best of my ability.
Consider the box of books to be a point mass at the origin. The student pulls with a force of 120N in the positive x direction (right). Since the box is on a surface, the Normal force (N) will be equal and opposite the Force due to gravity (Weight, W) The frictional force (f) acts against the force in the positive x direction, so this force goes left.
This can be shown mathematically by equations of sums of forces in particular directions.
∑Fy = N – W = 0 This equation says that the y direction is in static equilibrium meaning that the acceleration is 0. This equation confirms that the Normal force is equal to the weight. W = mg
thus…
N = mg
The frictional force is equal to the normal force times the coefficient of friction f = Nu = mgu
∑Fx = F – f = ma There is motion in this direction thus the sum of the forces equals the mass times the acceleration according to Newton’s 2nd law
F – mgu = ma
120 – (20kg)(9.81m/s/s)(.5) = (20kg)a
1.095 m/s/s = a
For part b the applied force is now at an angle and we need to incorporate the vector components of the force…there is still no motion in the y direction but the normal force will change slightly
∑Fy = N + Fsin30 – W = 0
N = W – Fsin30 = mg – 120sin30 = 20kg(9.81m/s/s) – 120(1/2) = 136.2
f = Nu = 136.2(.5) = 68.1N
The sum of the forces in the x direction will change slightly as well
∑Fx = Fcos30 – f = ma
103.923 – 68.1 = (20kg)a
35.82305N / 20kg = a
a = 1.79 m/s/s
2° HOMENAGEM AOS AMIGOS DO FACE BOOK (( 2° VERSÃO 2.2 AFB2011)).mpg
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