Please help me with question 16 and question 17 . Thank you

A mass of 3.0 kg rests on a smooth surface inclined 34° above the horizontal. It is kept from sliding down the plane by a spring

azamat

Answer:

The spring stretched by x = 13.7 cm

Explanation:

Given data

Mass = 3 kg

k = 120 $\frac{N}{m}$

Angle $\theta$ = 34°

From the free body diagram

Force acting on the box = mg sin $\theta$

⇒ F = 3 × 9.81 × $\sin34$

⇒ F = 16.45 N ------- (1)

Since box is attached with the spring so a spring force also acts on the box.

$F_{sp}$ = k x

$F_{sp}$ = 120 $x$ -------- (2)

The net force acting on the body is given by

$F_{net} = ma$

Since acceleration of the box is zero so

$F_{net} = 0$

$F - F_{sp} = 0$

$F = F_{sp}$

Put the values from equation (1) & (2) we get

16.45 = 120 $x$

x = 0.137 m

x = 13.7 cm

Therefore the spring stretched by x = 13.7 cm

3 0

2 years ago

Which force stops the car from moving?

Alex73 [517]

Answer:

The force of friction.

Explanation:

Gravity keeps the car on the ground.

Motion Allows the car to move.

The force of speed doesnt make sense.

Friction would cause the car to stop moving.

8 0

2 years ago

Read 2 more answers

A 50 kg runner runs up a flight of stairs. The runner starts out covering 3 steps every second. At the end the runner stops. Thi

nadezda [96]

To solve the problem it is necessary to take into account the concepts of kinematic equations of motion and the work done by a body.

In the case of work, we know that it is defined by,

$W = F * d$

Where,

F= Force

d = Distance

The distance in this case is a composition between number of steps and the height. Then,

$d=h*N$ , for h as the height of each step and N number of steps.

On the other hand we have the speed changes, depending on the displacement and acceleration (omitting time)

$V_f^2-V_i^2 = 2a\Delta X$

Where,

$V_f =$ Final velocity

$V_i =$ Initial Velocity

a = Acceleration

$\Delta X =$ Displacement

PART A) For the particular case of work we know then that,

$W = F*d$

$W = m*g*(h*N)$

$W = 50*9.8*(0.3*30)$

$W = 4.41kJ$

Therefore the Work to do that activity is 4.41kJ

PART B) To find the acceleration (from which we can later find the time) we start from the previously given equation,

$V_f^2-V_i^2 = 2a\Delta X$

Here,

$V_i = \frac{0.3*3}{1} = 0.90m/s\rightarrow$ 3 steps in one second

$v_f = 0$

Replacing,

$V_f^2-V_i^2 = 2a\Delta X$

$0-0.9^2=2a(30*0.3)$

Re-arrange for a,

$a = -\frac{0.9^2}{2*30*0.3}$

$a = -45*10^{-3}m/s^2$

At this point we can calculate the time, which is,

$t = \frac{\Delta V}{a}$

$t = \frac{0-0.9}{-45*10^{-3}}$

$t = 20s$

With time and work we can finally calculate the power

P = \frac{W}{t} = \frac{4.41}{20}

P = 0.2205kW

6 0

3 years ago

How long will it take a 2190 W motor to lift a 1.47 x 104 g box, 6.34 x 104 mm vertically.

Rasek [7]

Answer:

t = 4.17 [s]

Explanation:

We know that work is defined as the product of force by distance.

W = F*d

where:

F = force [N] (units of Newtons)

d = distance = 6.34 x 10⁴ [mm] = 63.4 [m]

In order to find the force, we must determine the weight of the box, the weight can be determined by means of the product of mass by gravitational acceleration.

w = m*g

where:

m = mass = 1.47 x 10⁴ [g] = 14.7 [kg]

g = gravity acceleration = 9.81 [m/s²]

w = 14.7*9.81

w = 144.2 [N]

Therefore the work can be calculated.

W = w*d

W = 144.2*63.4

W = 9142.72 [J] (units of Joules)

Power is now defined in physics as the relationship of work at a given time

P = W/t

where:

P = power = 2190 [W]

t = time [s]

Now clearing t, we have.

t = W/P

t = 9142.72/2190

t = 4.17 [s]

6 0

2 years ago

A coin is dropped. What is its displacement after 0.30 s.

Morgarella [4.7K]

Assuming this coin is on earth and that it wasn’t dropped forcefully:
Use the formula d = 1/2at^2. Rewriting using a=g and solving for height h gets us h = 1/2(9.8)t^2.
In this case that would get that the change in height h is 0.5(9.8)(0.3^2) = 0.441 m.

3 0

2 years ago