Question

An object whose weight is 100 lbf experiences a decrease in kinetic energy of 500 ft lbf and an increase in potential energy of 1500 ft lbf. The initial velocity and elevation of the object, each relative to the surface of the earth, are 40 ft/s and 30 ft, respectively. If g 5 32.2 ft/s2 , determine:
(a) the final velocity, in ft/s.
(b) the final elevation, in ft.

Answer 1

Answer:

a)  $v_2=35.60ft/sec$

b) $h_2=45ft$

Explanation:

From the question we are told that:

Weight $W=100lbf$

Decrease in kinetic energy $dK.E= 500 ft lbf$

Increase in potential energy $dP.E =1500 ft lbf.$

Velocity $V_1=40$

Elevation $h=30ft$

$g=32.2 ft/s2$

a)

Generally the equation for Change in Kinetic Energy is mathematically given by

$dK.E=\frac{1}{2}m(v_1^2-v_2^2)$

$500=\frac{1}{2}*\frac{100}{32.2}(v_1^2-v_2^2)$

$v_2=35.60ft/sec$

b)

Generally the equation for Change in Potential Energy is mathematically given by

$dP.E=mg(h_2-h_1)$

$1500=mg(h_2-h_1)$

$h_2=45ft$