Problem 1 Fnet,x = max Fnet,y = may fk = N vx^2 = v0x^2 +...

Problem 1 $F_{net,x} = ma_x$ $F_{net,y} = ma_y$ $f_k = \mu_k N$ $v_x^2 = v_{0x}^2 + 2a_x(x - x_0)$ (a) Show that the velocity of the box when it reaches the bottom of the ramp, ignoring friction, is 7 m/s using Newton's 2nd Law. (b) Show that the velocity of the box when it reaches the bottom of the ramp, including kinetic friction, $(\mu_k = 0.5)$ is 2.56 m/s using Newton's Second Law.

Transcript

00:01 So, here in this problem, we are given with this figure and it's a very interesting question.

00:06 So let me first explain what the question says.

00:08 So it says that a person wants to lift a 12 kg cart on this inclined plane and this length is 2 .5 meter it is given.

00:19 Now a worker calculates that if he gives the initial speed of 5 meter per second to the block, it will reach to the truck.

00:30 But he ignored friction while calculating the speed and he let it go with this much of speed.

00:36 But due to friction, this box slides only up to a distance of 1 .6 meter here.

00:43 It stops here and then slides back onto this plane.

00:49 Now there are two parts in this question that we need to solve.

00:53 In the a part, we have to find the magnitude of the friction force acting on the crate and also we have to assume that the friction remains constant.

01:04 So let us try to solve the a part first.

01:08 The mass of the crate is 12 kg given and the length up to which the crate moves is only 1 .6 meter.

01:17 It is already given in the question.

01:18 Now this angle is, the angle of this inclined plane is 30 degrees and the initial speed when it is moving up it is 5 meter per second and final speed at this point, the final speed becomes 0.

01:36 So our focus would be between these two points, point 1 and point number 2.

01:41 Now to find the friction force acting, we are going to use the work energy theorem.

01:51 It's a very powerful theorem and it's a general theorem which can be used anywhere in physics.

01:58 So let us use that theorem, the work energy theorem.

02:02 According to this, the total work done on the object is equal to the change in kinetic energy of the object.

02:11 So if you consider the forces acting on the block when it is moving up, first is the friction force which is acting in the backward direction f and the component of mg force in the backward direction and this component will be mg sin theta.

02:30 So only two forces are acting when it is moving up.

02:34 So the direction of the object is towards this side and both of these forces are acting backwards.

02:41 So work done would be negative.

02:42 So what i can write here that minus mg into sin theta multiplied with d minus friction multiplied with d.

02:54 This is the total work done is equal to the change in kinetic energy.

02:58 So final kinetic energy minus the initial kinetic energy.

03:05 Now at the second point, the final kinetic energy becomes zero.

03:09 So this term goes to zero because the speed becomes zero at second point.

03:13 So kinetic energy becomes zero.

03:16 Now from here, we're going to find the expression for f.

03:18 So minus f into d, this will be equals to minus one over two m v i square plus m into g into sin theta into d.

03:34 Now further f will be equals to this negative sign goes here.

03:38 So one over two m v i square minus mg into sin theta into d divided by d.

03:50 Correct? so this is expression for friction force.

03:52 Now just put the values to find the answer.

03:54 So f this will be equals to one over two.

03:58 Mass is given to be 12 kg.

04:00 V i is five.

04:02 So five square minus m is 12 kg.

04:06 G standard 9 .81.

04:08 Sin theta is 30 degrees...

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