A 2.8-kg block slides over the smooth, icy hill shown in...

A 2.8-kg block slides over the smooth, icy hill shown in Fig. P7.50. The top of the hill is horizontal and $70 \mathrm{~m}$ higher than its base. What minimum speed must the block have at the base of the hill in order for it to pass over the pit at the far side of the hill?

Key Concepts

Conservation of Energy

This principle states that in the absence of nonconservative forces, the total mechanical energy (the sum of kinetic and potential energy) remains constant throughout the motion. It is used to relate the speed of an object at different heights by equating the loss in gravitational potential energy to the gain in kinetic energy, or vice versa.

Gravitational Potential Energy

Gravitational potential energy is the energy stored in an object due to its elevation in a gravitational field. It plays a key role in analyzing vertical motion, as changes in height result in the conversion of potential energy into kinetic energy, which influences the speed of the object.

Kinetic Energy

Kinetic energy is the energy an object possesses by virtue of its motion. In dynamics problems, it is a crucial concept for determining the speed of the object when energy is converted from or to kinetic energy during its motion over a surface or through a system.

Centripetal Force Requirement in Curved Motion

When an object moves along a curved path, such as the crest of a hill, a centripetal force is necessary to keep it following the curved trajectory. The critical condition, often determined by setting the normal force to zero, establishes the minimum speed needed at the top of the curve to ensure that the object does not lose contact with the surface.

Transcript

00:01 Problem 746, a 2 .8 kilogram block slides over a smooth icy hill shown in figure 746.

00:07 The top of the hill is horizontal and it's 70 meters higher than its base.

00:14 What moon speed must block have at the base of this hill to pass over the pit? so the pit is right here.

00:20 So we're assuming that it's going to be taking a little parabolic arc as it goes down to this point here.

00:28 Okay, so to start off, this needs to be set up as a kinematic problem, then we'll get into conservation and energy later.

00:36 So we need to first find, i'm going to first find what the velocity is here at this point.

00:43 In order to do that, i need to find what the two pieces are.

00:46 There's v of x and there's v of y.

00:48 So in order to use both, i need to find how much time it takes to travel.

00:53 Okay, so in order to go from point, in order to go down from an elevation of a to b, that's going to be, uh, versus, is only going to be gravity acting in that dimension.

01:05 So i need to set up the distance is equal to 1 1 half a t squared.

01:11 So that's going to be 20 meters.

01:13 The difference between those two, 70 meters minus 50 meters, is equal to 1 half g t squared.

01:22 This is from the basic kinematic problem, the kinematic equation, where you would have the initial velocity plus the initial distance, plus the initial distance times time, velocity times time, plus the acceleration portion.

01:37 Okay, so then this gets us t is equal to the square root of 40 over g.

01:45 So that's going to be 40 meters divided by meters per second.

01:50 Okay, so then what we do now is we write vy is equal to gt, because stat is starting from zero.

02:02 It goes down for a certain amount of time at acceleration.

02:06 So then that's going to be the vertical velocity in the y direction at that point.

02:13 So then that's going to be a square root of 40g...

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