a student places a 2.00 Kg block on an inclined plane 3. A student places a 2.00

Question

a student places a 2.00 Kg block on an inclined plane 3. A student places a 2.00-kg block on an inclined plane, as shown in Figure 1 (a), and measures the block's acceleration to be 1.40 m/s. The student then lowers the inclined plane to make it horizontal, and ties the block to a string that passes over a frictionless pulley and connects to a 1.00-kg hanging mass. This new arrangement is shown in Figure 1 (b). Calculate the expected acceleration of the block in the new arrangement. Note that the same block and plane are used in both arrangements (with the latter changing its inclination angle). 2.00 kg 2.00 kg 70.0 1.00kg Figure 1

Explanation / Answer

along the incline,

Fnet = 2 g cos70 - f = 2 a

f = 2 (9.81 cos70 - 1.40) = 3.91 N

and f = u m g sin70

u = 3.91 / (2 x 9.81 x sin70) = 0.212

now on 1 kg :

g - T = a ..... (i)

on 2kg: T - (0.212 x 2 g ) = 2 a .... (ii)

(i) + (ii) => 9.81 - (0.424 x 9.81) = 3 a

a = 1.88 m/s^2 .......Ans

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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