A 18.0-kg cart is moving with a velocity of 7.30 m/s down a level hallway. A constant force of -13.0 N acts on the cart and its velocity becomes 3.30 m/s.
(a) What is the change in kinetic energy of the cart?
1 . J
(b) How much work was done on the cart?
2 J
(c) How far did the cart move while the force acted?
3 m
1) I did (1/2)18(7.30^2)=479.61
then (1/2)18(3.30^2)=98.01
479.61-98.01=381.6 (wrong?)
(a)ΔKE =KE2 – KE1 = m•v2²/2 - m•v1²/2 =
= m•(v2²- v1²)/2 =
= 18(3.3² - 7.3²)/2 = - 381.2 J.
(b) W = ΔKE = - 381.2 J.
(c) F = m•a,
a = F/m = - 13/18 = - 0.72 m/s².
s =(v² - vₒ²)/2•a =
={(3.3) -( 7.3)²)/2•(-0.72) = 29.44 m/s².
To find the change in kinetic energy of the cart, you correctly used the formula for kinetic energy: KE = (1/2)mv^2, where m is the mass and v is the velocity.
For the initial velocity, you used 7.30 m/s, and for the final velocity, you used 3.30 m/s. However, it seems that you made a calculation error when squaring the velocities.
Calculating the initial kinetic energy:
KE_initial = (1/2) * 18 kg * (7.30 m/s)^2
KE_initial = (1/2) * 18 kg * 53.29 m^2/s^2
KE_initial = 479.61 J
Calculating the final kinetic energy:
KE_final = (1/2) * 18 kg * (3.30 m/s)^2
KE_final = (1/2) * 18 kg * 10.89 m^2/s^2
KE_final = 98.01 J
To find the change in kinetic energy, subtract the final kinetic energy from the initial kinetic energy:
Change in KE = KE_final - KE_initial
Change in KE = 98.01 J - 479.61 J
Change in KE = -381.60 J (note the negative sign)
Therefore, the correct answer to (a) is that the change in kinetic energy of the cart is -381.60 J.
Let's proceed to solve parts (b) and (c) of the question.