A) constant friction force of 4.8 N retards the block\'s motion from the moment

Question

A) constant friction force of 4.8 N retards the block's motion from the moment it is released. Using an energy approach, find the position x of the block at which its speed is a maximum.

B) Explore the effect of an increased friction force of 14.5 N. At what position of the block does its

maximum speed occur in this situation??

A block of mass 1.5 kg is attached to a horizontal spring that has a force constant 1 100 N/m as shown in the figure below. The spring is compressed 2.0 cm and is then released from rest. constant friction force of 4.8 N retards the block's motion from the moment it is released. Using an energy approach, find the position x of the block at which its speed is a maximum. Explore the effect of an increased friction force of 14.5 N. At what position of the block does its maximum speed occur in this situation??

Explanation / Answer

Initial Energy
E = 1/2 k d^2

At maximum speed
E' = 1/2 k x^2 + 1/2 m^2

W = dE
F (d - x) = 1/2 k d^2 - (1/2 k x^2 + 1/2 m v^2)
v^2 = k d^2 / m - k x^2 / m - 2F (d - x) / m

differentiate and vmax if dv/dx = 0
2v dv/dx = - 2 k x / m + 2F / m = 0
F = k x
x = F / k

a)
x = 4.8 / 1100 = 4.36 x 10^-3 m = 0.436 cm

b)
x = 14.5 / 1100 = 0.0131818 m = 1.318 cm

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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