Recent content by HarshK

I think most of the collision was elastic but initially was inelastic for a very small distance.

So with this in mind, I'm guessing I will have to re-calculate the velocity of the box, the acceleration and coefficient value. m1v1+m2v2=(m1+m2)v1' (0.125)(0.71)+(0.2673)(0) = (0.125+0.2673)(v1'), v1' = 0.23m/s (vf^2-vi^2)/2d = a (0^2 - 0.23^2) / 2(0.032) = a, a =-0.83 m/s -0.83/9.8 = μ μ =...

So with this in mind, I'm guessing I will have to re-calculate the velocity of the box, the acceleration and coefficient value. m1v1+m2v2=(m1+m2)v1' (0.125)(0.71)+(0.2673)(0) = (0.125+0.2673)(v1'), v1' = 0.23m/s (vf^2-vi^2)/2d = a (0^2 - 0.23^2) / 2(0.032) = a, a =-0.83 m/s -0.83/9.8 = μ μ...

Wait I have a revision, the collision was inelastic. Asked a friend to confirm. So I believe haruspex may be right when he said m1v1+m2v2=(m1+m2)v1'

Me and my group made the measurements. The bob was released at an angle (30 degrees from rest / 60 degrees below x axis) and when released it hits the box, the box moves but the bob remains at the point where it was before moving it to an angle. So, like 90 degrees to the box after being hit...

Alright I'll be sure to post correctly next time! Thanks for your help, it helped a great load:smile:!

Ahh I see. So, the box stops at 0.032 meters so Vf = 0 but has an initial velocity of 0.33 m/s so Vi = 0.33 m/s Vf^2 = Vi^2 + 2ad 0^2 = 0.33^2 + 2 (a) (0.032) -0.33^2 = 2 a (0.032) 0.1089 = 2 a (0.032) ((0.1089 / 2) / 0.032) = a a = 1.70 m/s^2 Did I do the calculation process correctly? If so...

Agreed! Fn = mg, so the normal force equals (0.2673)(9.8) = Fn = 2.62 N μFn = ma, (Second Law) so (0.2673)(a) Then ma/mg = μ But how do I find the value of 'a'? The box moved 3.2 cm but way too quickly for me to record an accurate time.

I'm guessing by forces you mean forces such as driving force. So, W=FD. 1/2mv^2=W. (1/2mv^2) / d = F. (1/2(0.2673)(0.33^2)) / 0.032 = 0.455 N I'm not sure if this was a valid calculation, but if it is is the friction equal to 0.455N?

Well, if I take conservation of momentum (m1+v1=m2+v2) into account then, (0.125)(0.84)=(0.2673)(v2). v2 being the velocity of the box after the collision, I calculated to be 0.33m/s.

I was only able to figure out the final velocity of the bob before it hit the box through energy. PE=KE mgh=1/2mv^2 v = sqrt(2*g*h)

Homework Statement I am investigating a scenario where a pendulum with a bob attached is released from an angle and pushes a box to a certain distance. My goal is to find the coefficient of friction between the box and the surface it moved on. I have measurements for: - mass of the bob (125...