Download the App!

Get 24/7 study help with the Numerade app for iOS and Android! Enter your email for an invite.

(I) A 265 -kg load is lifted 23.0 m vertically with an acceler- ation $a = 0.150 g$ by a single cable. Determine $( a )$ the tension in the cable; $( b )$ the net work done on the load; assuming it started from rest.

Get the answer to your homework problem.

Try Numerade free for 7 days

Like

Report

8.22 $\mathrm{m} / \mathrm{s}$

Physics 101 Mechanics

Chapter 7

Work and Energy

Work

Kinetic Energy

Potential Energy

Energy Conservation

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

University of Sheffield

University of Winnipeg

Lectures

04:05

In physics, a conservative force is a force that is path-independent, meaning that the total work done along any path in the field is the same. In other words, the work is independent of the path taken. The only force considered in classical physics to be conservative is gravitation.

03:47

In physics, the kinetic energy of an object is the energy which it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body in decelerating from its current speed to a state of rest. The kinetic energy of a rotating object is the sum of the kinetic energies of the object's parts.

08:08

A $265-\mathrm{kg}$ load i…

04:58

(III) A $285-\mathrm{kg}$ …

04:01

(III) A 265-kg load is lif…

05:21

05:22

02:16

4.71 A block of mass $20.0…

04:40

Knowing that the tension i…

02:46

A block of mass $20.0 \mat…

03:35

A skier of mass $70.0 \mat…

02:05

A cable with $20.0 \mathrm…

So let's draw everybody diagram. We have forced tension going straight up and then going straight down his force of gravity. We're going to say that for party, the sum of forces in the war direction equals a and so we know that is going to be equal to 0.15 times the acceleration due to gravity on earth surface. Therefore, forced tension minus M g will be equal to a 0.15 g. At this point, we know that forced tension is going to be equal to 1.15 mg and two find force tension. We can simply say that this is going to be 1.15 times the mass of 265 kilograms times the acceleration due to gravity 9.8 meters per second squared and we have the force. Tension is 2.99 times 10 to the third Nunes. Now for part B, we know that the net work done is going to be equal to the net Force times cosa net force times d co sign of data. We know that this is going to be equal toe one, so this will simply be 10.15 times Majidi and we can actually solve. So Lynette work done is going to be 0.15 times 265 kilograms times 9.8 meters per second squared times the distance of 23.0 meters. And this is giving us 8.96 times 10 to the third Nunes. So this is the network done. Ah, this is the magnitude of thie tension force. And then, for part. See, they want us to find the work done by the cable. This will simply be the force of the tension times D co sign of Sarah degrees. Once again, this will be one, and this will be 2.99 times 10 to the third Nunes times d the distance of 23.0 meters. And so we find that the work done by the cable will be equal to 6.87 times 10 to the fourth Jules Brother, My apologies. Yeah, my, uh wrong units, my policies work isn't Jules forces and Noone's my apologies. So this would be a fine answer for sea. And then for part d, they want us to find the work done by gravity, so this would simply be equal to negative mg co sign if Ada, We know that this is again one and this will equal negative 265 kilograms times 9.8 meters per second squared times the distance of 23.0 meters. And we find that the work done by gravity is going to be equal to negative 5.97 times 10 to the fourth. Jules, that would be your final answer for party. And then finally for party, we're trying to find the final velocity. So we know that the net work done is going to be equal to the changing kinetic energy. So this will be equal to 1/2 times V final squared minus V initial squared. We know that in this case, the initial square is going to be equal zero. So essentially the final velocity will be equal to the square root of two times the network divided by the mass to the 1/2 power. And so this is going to be equal. Tio two times 8.96 times 10 to the third Jules, divided by the mass of 265 kilograms all to the 1/2 power. We find that the final velocity is equaling eight point 22 meters per second. So this will be our final answer for apart E. That is the end of the solution. Thank you for watching.

View More Answers From This Book

Find Another Textbook

03:19

Two strings of linear mass densities u and 9u are stretched under same tensi…

01:20

The density of copper is 8.4 g/cm3. What will be the volume of 200 g of copp…

Find the mass of 50 cm3 of silver if its density is 10.5 g/cm3. Explain step…

01:13

20119.3.The position x of a particle varies with time t asx …

01:32

10. In your study room a plane mirror is fixed to the wall in front of you a…

02:31

An object 5cm high is placed from a convex mirror of radius of curvature 30c…

03:00

The weight of the ladder leaning at the wall produces an anticlockwise torqu…

01:17

25 kg of sand is deposited each second on a conveyor belt moving at 10m/s. T…

01:02

The amplitude aur an oscillating pendulum is 2.5 cm what is the distance bet…

01:22

Gold has a density of 19300 kg/m³. Calculate the mass of 0.02 m³ of gold in …