/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q55P  A 0.5 kg teddy bear is nudge... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A0.5kg teddy bear is nudged off a window sill and falls 2m to the ground. (a) What is the kinetic energy at the instant it hits the ground? What is its speed? What assumptions or approximations did you make in this calculation?

(b) A 1.0kg flowerpot is nudged off a window sill and falls 2m to the ground. What is the kinetic energy at the instant it hits the ground? What is its speed? How do the speed and kinetic energy compare to that of the teddy bear in part (a)?

Short Answer

Expert verified

(a) For 0.5kg teddy bear,

The speed is6.26m/s

The kinetic energy at the instant it hits the ground is 9.80J

(b) For1.0kg flowerpot

The speed is 6.26m/s

The kinetic energy at the instant it hits the ground is 19.59J

Step by step solution

01

Identification of given data

- The mass of a teddy bear is 0.5kg

- The height from which the teddy bear falls is 2m

- The mass of a flowerpot is 1.0kg

- The height from which the flowerpot falls is 2m

02

concept of kinetic energy

The kinetic energy is determined when an object is in motion. When the total force is applied on an object, the work done speeds up the object and gets kinetic energy.

03

(a) Calculation of Speed and kinetic energy of a teddy bear

The speed is calculated by the following formula,

v2=2gH

where,

g=acceleration due to gravity =9.81m/s2

H=height from which the teddy bear falls

Substitute the values in the above expression,

v=2gH

=2×9.81×2

=6.26m/s

Hence, the speed of a teddy bear is 6.26m/s

The kinetic energy is calculated by the following formula,

E=12mv2

where,

m=mass of a teddy bear

v-speed of a teddy bear

Substitute the values in the above expression,

E=12×0.5×6.262

=9.80J

Hence, the kinetic energy at the instant the bear hits the ground is 9.80J

The approximation and assumption made in this calculation is the "g" value which is due to the acceleration of gravity.

04

(b) Calculation of Speed and kinetic energy of a flowerpot

The speed is calculated by the following formula,

v2=2gH

where,

g=acceleration due to gravity =9.81m/s2

H=height from which the teddy bear falls

Substitute the values in the above expression,

v=2gH

=2×9.81×2

=6.26m/s

Hence, the speed of a teddy bear is 6.26m/s

The kinetic energy is calculated by the following formula,

E=12mv2

Where,

m=mass of a teddy bear

v-speed of a teddy bear

E=12×1×(6.26)2

=19.59J

Hence, the kinetic energy at the instant the bear hits the ground is 19.59J

The speed of a flowerpot and the bear is the same because both of them fall from the same height H.

The kinetic energy of a flower pot is twice of that of a teddy bear. Because the mass of the flowerpot is twice as the mass of a teddy bear.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

You push a heavy crate out of a carpeted room and down a hallway with a waxed linoleum floor. While pushing the crate 2.3mout of the room you exert a force of 30N; while pushing it 8mdown the hallway you exert a force of 15N. How much work do you do in all?

The radius of an airless planet is 2000km(2×106m) and its mass is 1.2×1023kg. An object is launched straight up from just above the atmosphere of Mars (a) What initial speed is needed so that when the object is far from the planet its final speed is 900m/s ? (b) What initial speed is needed so that when the object is far from the planet its final speed is 0m/s ? (This is called the escape speed.)

Question: The radius of the Moon is 1750 km, and its mass is 7×1022kg. What would be the escape speed from an isolated Moon? Why was a small rocket adequate to lift the lunar astronauts back up from the surface of the Moon?

A constant force 23,12,32Nacts through a displacement0.12,0.31,0.24m. How much work does this force do?

You throw a ball of mass 1.2kgstraight up. You observe that it takes3.1sto go up and down, returning to your hand. Assuming we can neglect air resistance, the time it takes to go up the top is half the total time, 1.55 s . Note that at the top the momentum is momentarily zero, as it changes from heading upward to heading downward. (a) Use the momentum principle to determine the speed that the ball had just after it left your hand. (b) Use the Energy Principle to determine the maximum height above your hand reached by the ball.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.