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WARRIOR [948]
1 year ago
13

The slider of mass m is released from rest in position A and slides without friction along the vertical-plane guide shown. Deter

mine the height h such that the normal force exerted by the guide on the slider is zero as the slider passes point C. For this value of h, determine the normal force as the slider passes point B.
Physics
1 answer:
Anuta_ua [19.1K]1 year ago
8 0

The value of normal force as the slider passes point B is

  • 6 mg

The value of h when the normal force is zero

  • 3R/2

<h3>How to solve for the normal force</h3>

The normal force is calculated using the work energy principle which is applied as below

K₁ + U₁ = K₂

k represents kinetic energy

U represents potential energy

the subscripts 1,2 , and 3 = a, b, and c

for 1 to 2

K₁ + W₁ = K₂

0 + mg(h + R) = 0.5mv²₂

g(h + R) = 0.5v²₂

v²₂ = 2g(1.5R + R)

v²₂ = 2g(2.5R)

v²₂ = 5gR

Using summation of forces at B

Normal force, N  = ma + mg

N = m(a + g)

N = m(v²₂/R + g)

N = m(5gR/R + g)

N = 6mg

for 1 to 3

K₁ + W₁ = K₃ + W₃

0 + mgh = 0.5mv²₃ + mgR

gh = 0.5v²₃ + gR

0.5v²₃ = gh - gR

v²₃ = 2g(h - R)

at C

for normal force to be zero

ma = mg

v²₃/R = g

v²₃ = gR

and v²₃ = 2g(h - R)

gR = 2gh - 2gR

gR + 2gR = 2gh

3gR = 2gh

3R/2 = h

Learn more about normal force at:

brainly.com/question/20432136

#SPJ1

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2. A = 0.4 m/s^2

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Explanation:

3 0
3 years ago
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Running at 1.55 m/s, Bruce, the 40.0 kg quarterback, collides with Biff, the 90.0 kg tackle, who is traveling at 7.0 m/s in the
Margarita [4]

Answer:Bruce is knocked backwards at  

14

m

s

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Explanation:

This is a problem of momentum (

→

p

) conservation, where

→

p

=

m

→

v

and because momentum is always conserved, in a collision:

→

p

f

=

→

p

i

We are given that  

m

1

=

45

k

g

,  

v

1

=

2

m

s

,  

m

2

=

90

k

g

, and  

v

2

=

7

m

s

The momentum of Bruce (

m

1

) before the collision is given by

→

p

1

=

m

1

v

1

→

p

1

=

(

45

k

g

)

(

2

m

s

)

→

p

1

=

90

k

g

m

s

Similarly, the momentum of Biff (

m

2

) before the collision is given by

→

p

2

=

(

90

k

g

)

(

7

m

s

)

=

630

k

g

m

s

The total linear momentum before the collision is the sum of the momentums of each of the football players.

→

P

=

→

p

t

o

t

=

∑

→

p

→

P

i

=

→

p

1

+

→

p

2

→

P

i

=

90

k

g

m

s

+

630

k

g

m

s

=

720

k

g

m

s

Because momentum is conserved, we know that given a momentum of  

720

k

g

m

s

before the collision, the momentum after the collision will also be  

720

k

g

m

s

. We are given the final velocity of Biff (

v

2

=

1

m

s

) and asked to find the final velocity of Bruce.

→

P

f

=

→

p

1

f

+

→

p

2

f

→

P

f

=

m

1

v

1

f

+

m

2

v

2

f

Solve for  

v

1

:

v

1

f

=

→

P

f

−

m

2

v

2

f

m

1

Using our known values:

v

1

f

=

720

k

g

m

s

−

(

90

k

g

)

(

1

m

s

)

45

k

g

v

1

f

=

14

m

s

∴

Bruce is knocked backwards at  

14

m

s

.

Explanation:

5 0
2 years ago
What is the volume of an object if it has a mass of 10 grams and a density of 87 g/ml
belka [17]

Answer:

<h3>The answer is 0.115 mL</h3>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

volume =  \frac{mass}{density} \\

From the question

mass = 10 g

density = 87 g/ml

We have

volume =  \frac{10}{87}  \\  = 0.114942528...

We have the final answer as

<h3>0.115 mL</h3>

Hope this helps you

5 0
3 years ago
You have a spring that stretches 0.070 m when a 0.10-kg block is attached to and hangs from it at position y0. Imagine that you
olga nikolaevna [1]

Answer:

a) Δy = 0.144 m

b) W = 0.145 J

c) Us = 0.32 J

d) ymax = 0.144 m

Explanation:

a) First let's find the spring constant using Hooke's Law

F = k*Δy   ⇒  k = F/Δy

where

F = m*g = 0.1 kg*9.81 m/s² = 0.981 N

and  Δy = 0.07 m. Hence

k = 0.981 N/0.07 m = 14.014 N/m ≈ 14 N/m

In order to find the position of the block when we let it go, we need to find the force that caused this expansion in the spring, we know that the reading of the scale was 3 N and this reading includes the force we want to find and the weight of the block, therefore:

f = 3 N - F = 3 N - 0.981 N = 2.019 N

Now that we have found the force we can use Hooke's Law in order to find the position of the block

f = k*Δy   ⇒   Δy = f/k

⇒   Δy = 2.019 N/14 N/m

⇒   Δy = 0.144 m

b) First, notice that there are two kind of potential energy: the potential energy in the spring and the potential energy due to the gravitational field:

W = ΔU

W = ΔUs + ΔUg

W = (Usf - Usi) + (Ugf - Ugi)

Notice that

Us = 0.5*k*y²

where

yf = 0.07 m + 0.144 m = 0.214 m  and

yi = 0.07 m

and we will take the zero level to be the equilibrium position where the block was hanging at rest. Hence

W = 0.5*k*(yf² - yi²) + m*g*(0 - Δy)

⇒ W = 0.5*14 N/m*((0.214 m)² - (0.07 m)²) + (0.1 kg)*(9.81 m/s²)*(0 - 0.144 m)

⇒ W = 0.145 J

c) When we let the block go the spring was stretched by

y = 0.07 m + 0.144 m = 0.214 m

Therefore:

Us = 0.5*k*y²

⇒ Us = 0.5*14 N/m*(0.214 m)²

⇒ Us = 0.32 J

d) Because the position that we pulled the block to it is considered as the amplitude for the vibrational motion that will happen after we release the block, then the maximum height the particle will reach above the equilibrium position is

ymax = Δy = 0.144 m

 

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