Physics

Question

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 a force of 280 N at an angle of 35° above the horizontal.

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

  • 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 re-write this as

    [tex]R=mg-Fsin \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 (mg-F 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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  • Answer:

    3.53 m/s^2

    Explanation:

    Use the net x value.

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