I\'ve answered (a) and (b) and they are correct, but I\'m having trouble finding

Question

I've answered (a) and (b) and they are correct, but I'm having trouble finding (c). Please explain how you got the answer, thank you.
A 7 kg box with initial speed 6 m/s slides across the floor and comes to a stop after 1.7 s.

(a) What is the coefficient of kinetic friction?
µk = .36

(b) How far does the box move?
distance = 5.1 m

(c) You put a 3 kg block in the box, so the total mass is now 10 kg, and you launch this heavier box with an initial speed of 6 m/s. How long does it take to stop?
?t = s

Explanation / Answer

For part c, the coefficient of kinetic friction is still the same, but the normal force has changed (since mass increased). You need to find the acceleration the box will have: To get this: Normal force = 10kg * 9.8 = 98 N Force from kinetic friction = uK * Normal Force = 0.36 * 98 N = 35.28 N Now, since F = m*a, you know: 35.28 N = 10kg * a a = 3.528 m/s^2 Finally, with acceleration, initial velocity, and final velocity known, you can find the time it takes with the equation: Vf = Vi + a * t 0 m/s = 6 m/s + 3.528 m/s^2 * t t = 1.70 seconds So it takes 1.70 seconds to come to a stop.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---