April works at the grocery store. She needs to move a heavy crate of bananas (m = 65 kg) to the produce section. She drags the crate along the floor by applying
Question
a. If there is no friction, what is the acceleration of the crate? (5 points)
280/65=4.3
b. If the coefficient of kinetic friction between the crate and the floor is 0.4, what is the acceleration of the crate? (5 points)
2 Answer

1. User Answers skyluke89
a) The acceleration in absence of friction is [tex]3.5 m/s^2[/tex]
b) The acceleration in presence of friction is [tex]0.6 m/s^2[/tex]
Explanation:
a)
First of all, we notice that the crate accelerates only along the horizontal direction, since the forces in the vertical direction are balanced (in fact, the weight is balanced by the normal reaction + the vertical component of the applied force).
This means that we can consider only the horizontal component of the applied force, which is
[tex]F_x = F cos \theta = (280)(cos 35^{\circ})=229.4 N[/tex]
where
F = 280 N is the applied force
[tex]\theta=35^{\circ}[/tex] is the angle
Now we can find the acceleration of the crate, by using Newton's second law:
[tex]F_x = ma[/tex]
where
[tex]F_x = 229.4 N[/tex] is the net force along the horizontal direction
m = 65 kg is the mass of the crate
a is the acceleration
Solving for a, we find
[tex]a=\frac{F_x}{m}=\frac{229.4}[65}=3.5 m/s^2[/tex]
b)
In this case, we also have the force of friction, so we have to compute the force of friction.
We start by writing the equation of the forces along the vertical direction:
[tex]R+F sin \theta  mg = 0[/tex]
where
R is the normal reaction
[tex]F sin \theta[/tex] is the vertical component of the applied force
[tex]mg[/tex] is the weight
We can rewrite this as
[tex]R=mgFsin \theta[/tex] (1)
Now we can write the equation of the forces along the horizontal direction
[tex]F cos \theta  \mu_k R = ma[/tex] (2)
where
[tex]F cos \theta[/tex] is the horizontal component of the applied force
[tex]\mu_K R[/tex] is the force of friction, with [tex]\mu_k = 0.4[/tex] is the coefficient of friction
By substituting (1) into (2) and solving for a, we find the new acceleration of the crate:
[tex]F cos \theta  \mu_k (mgF sin \theta) = ma\\a=\frac{F cos \theta  \mu_k mg +\mu_k F sin \theta}{m}=\frac{(280)(cos 35)(0.4)(65)(9.8)+(0.4)(280)(sin 35)}{65}=0.6 m/s^2[/tex]
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2. User Answers snoopy1420
Answer:
3.53 m/s^2
Explanation:
Use the net x value.