PHYSICS COLLEGE

across a rough, horizontal surface. The chair's mass is 18.8 kg. The force you exert on the chair is 165 N directed 26 degrees below the horizontal. While you slide the chair a distance of 6.00 m , the chair's speed changes from 1.30 m/s to 2.50 m/s . Find the work done by friction on the chair.

Answers

Answer 1

Answer:

-847.2J

Explanation:

First find the acceleration from v^2= u^2 + 2as

v= 2.5 m/s

u= 1.3 m/s

a???

s=6.00

a= v^2-u^2/2s

a= (2.5)^2-(1.3)^2/2× 6

a= 0.38ms^-2

From Newtons second law:

(Force applied cos Θ) - (Frictional force) = ma

Frictional force = ma- (Force applied cos Θ)

Frictional force= (18.8×0.38) - (165 cos 26°)

Frictional force= 7.144- 148.3= -141.2N

Therefore,

Work done by friction = Frictional force × distance covered

= -141.2N × 6= -847.2J

Answer 2

Answer:

W = –847J

Explanation:

Given m = 18.8kg, F = 165N, θ = -26° (below the horizontal, s = 6.0m, u = 1.30m/s and v = 2.50m/s

In this problem, two forces act on the chair; the forward force F and the frictional force f. We would apply newton's second law to find the frictional force f after which we can calculate the workdone by the frictional force f×s.

But for us to apply newton's second law, we need to know the acceleration of the chair cause by the net force.

From constant acceleration motion equations

v² = u² + 2as

2.5² = 1.30² + 2a×6

6.25 = 1.69 +12a

12a = 6.25 – 1.69

12a = 4.56a

a = 4.56/12

a = 0.38m/s

By newton's second law the net sum of forces equals m×a

The force F has horizontal and vertical and components. It is the horizontal component of this force that pushes the chair against friction.

Fx and f are oppositely directed.

Fx – f =ma

165cos(-26) – f = 18.8×0.38

148.3 – f = 7.14

f = 148.3 – 7.14

f = 141.2N

Workdone = -fs = –141.2×6.00 = –847J

W = –847J

Work is negative because it is done by a force acting on the chair in a direction opposite (antiparallel) to that of the intended motion.

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A helium balloon ride lifts up passengers in a basket. Assume the density of air is 1.28 kg/m3 and the density of helium in the balloon is 0.18 kg/m3. The radius of the balloon (when filled) is R = 4.9 m. The total mass of the empty balloon and basket is mb = 121 kg and the total volume is Vb = 0.073 m3. Assume the average person that gets into the balloon has a mass mp = 73 kg and volume Vp = 0.077 m3. 1)What is the volume of helium in the balloon when fully inflated? m3 2)What is the magnitude of the force of gravity on the entire system (but with no people)? Include the mass of the balloon, basket, and helium. N

Answers

Answer:

1) The volume of helium in the ballon when is fully inflated is 492.8070 m³

2) The magnitude of the force of gravity (with no people) is 869.3119 N

Explanation:

Given:

ρair = density of air = 1.28 kg/m³

ρhelium = density of helium = 0.18 kg/m³

R = radius of balloon = 4.9 m

mtotal = 121 kg

Vtotal = 0.073 m³

mp = average mass per person = 73 kg

Vp = 0.077 m³

g = gravity = 9.8 m/s²

Questions:

1) What is the volume of helium in the balloon when fully inflated, Vhelium = ?

2) What is the magnitude of the force of gravity on the entire system (but with no people), Fg = ?

1) The volume of helium in the ballon when is fully inflated

$V_(helium) =(4)/(3) \pi R^(3) =(4)/(3) \pi *4.9^(3) =492.8070m^(3)$

2) First, you need to calculate the mass of helium

$m_(helium) =\rho _(helium) *V_(helium) =0.18*492.8070=88.7053kg$

The magnitude of the force of gravity (with no people)

$F_(g) =m_(helium) *g=88.7053*9.8=869.3119N$

If a freely falling object were somehow equipped with a speedometer, its speedreading would increase each second by
a) about 15 m/s.
b) a rate that depends on its initial speed.
c) a variable amount.
d) about 5 m/s.
e) about 10 m/s.

Answers

Answer:

e) about 10 m/s

Explanation:

Acceleration due to gravity is nominally* 9.8 m/s². That means the change in velocity each second is ...

(9.8 m/s²)(1 s) = 9.8 m/s ≈ 10 m/s

_____

* This is the value expected to be used in the solution of many math and physics problems. The standard value of 'g' on Earth is defined as 9.80665 m/s². It varies from place to place and with altitude. At any given place, it may also vary with time as a result of changes in mass distribution within the Earth.

A 84-kg man stands on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.2 m/s in 0.73 as. The elevator travels with this constant speed for 5.0 s, undergoes a uniform negative acceleration for 1.4 s, and then comes to rest.What does the spring scale register During the first 0.80s of the elevator’s ascent?

Answers

Answer:

$SR=949.2N$

Explanation:

From the question we are told that:

Mass $M=84kg$

Speed $V=1.2m/s$

Acceleration Time $t_a=0.73$

Constant speed Time $t_s=5.0s$

Deceleration time $t_d=1.4s$

Generally the equation for Acceleration is mathematically given by

$a=(v)/(t)$

Therefore acceleration for the first 0.80 sec is

$a=(1.2)/(0.80)$

$a=1.5m/s^2$

Therefore

Spring Reading=Normal force -Reaction

$SR=m(g+a)$

$SR=84(9.8+1.5)$

$SR=949.2N$

A train travels due south at 20 m/s. It reverses its direction and travels due north at 20 m/s. What is the change in velocity of the train? 50 m/s, due south e50 m/s, due north 120 m/s, due south zero ms 40 m/s, due north

Answers

Answer:

40 m/s due north

Explanation:

Consider that the south direction a negative Y axis and north direction as + Y axis

v1 = 20 m/s South = 20 (-j) m/s

v2 = 20 m/s North = 20 j m/s

Change in velocity = v2 - v1 = 20 j - 20 (-j) = 40 j m/s

So, change in velocity is 40 m/s due north.

How much electrical energy is used by a 400 W toaster that is operating for 5minutes?
A. 2000 J
B. 75,000 J
C. 120,000 J
D. 300,000 J

Answers

The electrical energy used by a 400 W toaster that is operating for 5 minutes will be 120,000 J.Option C is correct.

What is the power output?

The rate of the work done is called the power output. It is denoted by P.Its unit of a watt. It is the ratio of the work done or the enrgy to the time period.

The given data in the problem is;

E is the electrical energy

P is the power output = 400 W

t is the time period = 5 minutes

The power output is given as;

$\rm P= (E)/(t) \n\n\ E= P * t \n\n\ E= 400 * 300 \n\n\ E=120,000 \ J$