Question

The 10-lb block has a speed of 4 ft>s when the force of F = (8t2) lb is applied. Determine the velocity of the block when it moves s = 30 ft. The coefficient of kinetic friction at the surface is ms = 0.2.

Answer 1

Answer:

The velocity of the block is 231650.8 ft/s

Explanation:

Knowing

W = 10 lb

v0 = u = 4 ft/s

F = 8 $t^{2}$ lb

ms = 0.2

Applying the newton second law

ΣFy = 0

N - W = 0 --> N = W = 10 lb

ΣFx = m$a_{x}$

F - ms N = ma

$8t^{2}$- 0.2(10) = $\frac{10}{32.2} a$

a = $3.22 (8t^{2} - 2)$

Using Kinematics

a = dv/dt

$\int\limits^v_u dv$ = $\int\limits^t_0 {a} \, dt$ = $\int\limits^t_o {3.22 (8t^{2} - 2)} \, dt$

v - 4 = 3.22 $(\frac{8}{2} t^{3} - 2t)$

when t = 30

v = 231650.8 ft/s