A block of mass m = 2.00 kg rests on the left edge of a block of mass M = 8.00 k

Question

A block of mass m = 2.00 kg rests on the left edge of a block of mass M = 8.00 kg. The coefficient of kinetic friction between the two blocks is 0.315 and the surface on which the 8.00-kg block rests is frictionless. A constant horizontal force of magnitude F = 12.5 N is applied to the 2.00-kg block, setting it in motion as shown in figure (a). If the distance L that the leading edge of the smaller block travels on the larger block is 3.00 m. In what time interval will the smaller block make it to the right side of the 8.00-kg block as shown in figure (b)? (Note: Both blocks are set into motion when is applied.) How far does the 8.00-kg block move in the process?

Explanation / Answer

Frictional force of the smaller block Fs = us mg

Fs = 0.315 * 2* 9.81

Fs = 6.18 N

accletation of the smaller block = as = F/m = 6.183/2 = 3.1 m/s^2

accleration of the larger block = am = F/M = F-Fs/M

aM = (12.5 - 6.18)/8 = 0.79 m/s^2

time interval for the smaller block = t^2 = (2L/a1-a2)

t^2 = (2 * 3)/(3.1-0.79)

t = 1.61 secs

-----------------

Distance travellerd S = 0.5 at^2

S = 0.5 * 0.79 * 1.61* 1.61

S = 1.023 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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