A student decides to move a box of books into her dormitory room by pulling on a

Question

A student decides to move a box of books into her dormitory room by pulling on a rope attached to the box. She pulls with a force of 82.4 N at an angle of 30.0° above the horizontal. The box has a mass of 24.0 kg, and the coefficient of kinetic friction between box and floor is 0.300. (Indicate the direction with the sign of your answer.) (a) Find the acceleration of the box. (Assume that the +x-axis is to the right and the +y-axis is up along the page.) magnitude direction m/s2 | ° counterclockwise from the +x-axis (b) The student now starts moving the box up a 10.0% incline, keeping her 82.4 N force directed at 3D O above the line of the incline.. If the coefficient of friction is unchanged, what is the new acceleration of the box? (Assume that the +x-axis is along the incline and the +y-axis is upwards and perpendicular to the incline.) magnitude direction m/s2 o counterclockwise from the incline

Explanation / Answer

Given mass m = 24 kg

coefficient of friction mue = 0.3

applied force F = 82 .4 N

theta = 30 deg

a) the net force acting on the box is

(F * cos(theta)) - f = m * a

(F * cos(theta)) - (mue * m * g) = m * a

(82.4 * cos(30)) - (0.3 * 24 * 9.8) = 24 * a

acceleration a = 0.033 m/s^2

the direction is along + x-axis

angle = 0 deg

b)

The force up the incline is F * cos(30)

the forces down the incline are

i) frictional force f = mue * m * g * cos(theta)

f = mue * m * g * cos(10)

ii) component of weight w = m * g * sin(10)

the net force is

(F * cos(30)) - (mue * m * g * cos(10)) - (m * g * sin(10)) = m * a

(82.4 * cos(30)) - (0.3 * 24 * 9.8 * cos(10)) - (24 * 9.8 * sin(10)) = 24 * a

a = - 1.62 m/s^2

angle is 180 deg ( down the incline)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.