All categories

4 years ago

A force does work on an object if a component of the force is

Physics

1 answer:

IrinaK [193]4 years ago

7 0

Answer:

A force does work on an object if a component of the force is parallel to the displacement of the object.

Explanation:

Work, a measurement of energy is said to be done when a force applied to an object results in the movement of that object to a certain distance and direction. Force is the act of push or pulls occurs on an object as a result of the interaction between that object with another one and displacement is the distance and direction covered by that object as a result of the force applied on it.

The work done (W) by a constant force (F) is equal to the product of the force in the direction of displacement of the object and the distance (d) moved by the object i.e., W = F * d.

The angle between the displacement and the force is θ, then the work done, W = Fd cos θ ........ (1)

Positive work - Force acts in the same direction with respect to the displacement of the object. Here, θ is zero, so cos θ i.e., cos 0 is 1. Therefore, from the equation (1), W = Fd (i.e., work done by the force is positive).

Negative work - Force acts in the opposite direction with respect to the displacement of the object. Here, θ is 180°, so cos θ i.e., cos 180° is -1. Therefore, from the equation (1), W = -Fd (i.e., work done by the force is negative).

If a force is applied to an object and it does not move, then the work done is zero i.e., W = F * 0 = 0. Also, if the force and displacement are at right angle to each other, then θ is 90°. Therefore, from the equation (1), W = 0 since cos 90° is zero.

You might be interested in

Create a list of three questions that are good science questions.

masya89 [10]

Answer:

what is speed

how to tell what speed an objeckt is moveing

vlocity

Explanation:

5 0

4 years ago

Based on the tug of war pictured below, what is the net force? –190 N 10 N –10 N 190 N.

umka21 [38]

Force is defined as the push or pull applied to the body. Its unit is Newton. The net force will be 10 N on the right side.

What is force?

Force is defined as the push or pull applied to the body. Sometimes it is used to change the shape, size, and direction of the body. Force is defined as the product of mass and acceleration. Its unit is Newton.

The given data in the problem is;

F₁ is forced by person 1 = 100 N

F₂ is forced by person 2= 90 N

Fₓ is the net force

Fₓ = F₁ -F₂

Fₓ = 100-90

Fₓ= 10 N

Hence the net force will be 10 N on the right side.

To learn more about the force refer to the link;

brainly.com/question/26115859

7 0

3 years ago

cycling at 13.0 m/s on your new aero carbon fiber bike, how much times would it take a pro cyclist to ride in 120 km? Give your

sp2606 [1]

We have that the time in seconds, minutes, and hours is

$t=1.083*10^{-4}s$

$T_{min}=1.805*10^{-6}min$

$T_{hours}=3.0083*10^{-8}hours$

From the Question we are told that

Velocity $v=13.0m/s$

Distance $d=120 km$

Generally the equation for the Time is mathematically given as

$t=\frac{13}{120*10^3}\\\\t=1.083*10^{-4}s$

Therefore

$T_{min}=1.083*10^{-4}s/60$

$T_{min}=1.805*10^{-6}min$

And

$T_{hours}=T_{min}/60$

$T_{hours}=3.0083*10^{-8}hours$

For more information on this visit

brainly.com/question/12319416?referrer=searchResults

4 0

3 years ago

A 5.0 g coin is placed 15 cm from the center of a turntable. The coin has static and kinetic coefficients of friction with the t

Alexandra [31]

Answer:

the coin does not slide off

Explanation:

mass (m) = 5 g = 0.005 kg

distance (r) = 15 cm = 0.15 m

static coefficient of friction (μs) = 0.8

kinetic coefficient of friction (μk) = 0.5

speed (f) = 60 rpm

acceleration due to gravity (g) = 9.8 m/s^{2}

lets first find the angular speed of the table

ω = 2πf

ω = 2 x π x 60 x $\frac{1}{60}$

ω = 6.3 s^{-1]

Now lets find the maximum static force between the coin and the table so we can get the maximum velocity the coin can handle without sliding

static force (Fs) = ma

static force (Fs) = μs x Fn = μs x m x g

Fs = 0.8 x 0.005 x 9.8 = 0.0392 N

Fs = ma

0.0392 = 0.005 x a

a = 7.84 m/s^{2}

$(Vmax)^{2}$ = a x r

$(Vmax)^{2}$ = 7.84 x 0.15

Vmax = 1.08 m/s

ωmax = $\frac{Vmax}{r}$

ωmax = $\frac{1.08}{0.15}$ = 7.2 s^{-1}

now that we have the maximum angular acceleration of the table, we can calculate its maximum speed in rpm

Fmax = $\frac{ωmax}{2π}$

Fmax = $\frac{7.2}{2 x π}$ = 68.7 rpm

since the table is rotating at a speed less than the maximum speed that the static friction can hold coin on the table with, the coin would not slide off.

4 0

4 years ago

A brass alloy is known to have a yield strength of 240 MPa (35,000 psi), a tensile strength of 310 MPa (45,000 psi), and an elas

Karo-lina-s [1.5K]

Answer:

Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.

Explanation:

Given that

Yield strength ,Sy= 240 MPa

Tensile strength = 310 MPa

Elastic modulus ,E= 110 GPa

L=380 mm

ΔL = 1.9 mm

Lets find strain:

Case 1 :

Strain due to elongation (testing)

ε = ΔL/L

ε = 1.9/380

ε = 0.005

Case 2 :

Strain due to yielding

$\varepsilon' =\dfrac{S_y}{E}$

$\varepsilon' =\dfrac{240}{110\times 1000}$

ε '=0.0021

Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.

For computation of load strain due to testing should be less than the strain due to yielding.

4 0

4 years ago