Question

A student moves a box of books down the hall by pulling on a rope attached to the box. The student pulls with a force of 187 N at an angle of 25.6 ◦ above the horizontal. The box has a mass of 31.5 kg, and µk between the box and the floor is 0.19. The acceleration of gravity is 9.81 m/s 2 . Find the acceleration of the box. Answer in units of m/s 2 . 019 (part 2 of 2) 10.0 points Now the student moves the box up a ramp (with the same coefficient of friction) inclined at 14.3 ◦ with the horizontal. b) If the box starts from rest at the bottom of the ramp and is pulled at an angle of 25.6 ◦ with respect to the incline and with the same 187 N force, what is the acceleration up the ramp? Answer in units of m/s 2 .

Answer 1

Answer:

Part a)

$a = 3.98 m/s^2$

Part b)

$a = 1.61 m/s^2$

Explanation:

Part a)

When box is pulled on rough horizontal plane then we will have

$F cos\theta - F_k = ma$

also we have

$F_n + Fsin\theta = mg$

$F_n = mg - Fsin\theta$

$F_n = (31.5 \times 9.81) - 187 sin25.6$

$F_n = 228.2 N$

Now we have

$F_k = \mu_k F_n$

$F_k = (0.19)(228.2)$

$F_k = 43.36 N$

Now we know that

$187 cos25.6 - 43.36 = 31.5 a$

$a = 3.98 m/s^2$

Part b)

Now When box is pulled on rough inclined plane then we will have

$F cos\theta - F_k - mgsin\phi= ma$

also we have

$F_n + Fsin\theta = mg cos\phi$

$F_n = mgcos\phi - Fsin\theta$

$F_n = (31.5 \times 9.81)cos14.3 - 187 sin25.6$

$F_n = 218.6 N$

Now we have

$F_k = \mu_k F_n$

$F_k = (0.19)(218.6)$

$F_k = 41.54 N$

Now we know that

$187 cos25.6 - 41.54 - (31.5\times 9.81 sin14.3) = 31.5 a$

$a = 1.61 m/s^2$