iSoul In the beginning is reality

Problems in mechanics, part 3

This post continues a series of problems (part 1 here, part 2 here) based on the website Free Solved Physics Problems, this time concentrating on dynamics problems for time-space corresponding to problems in space-time.

Note: A newton is how much force is required to make a mass of one kilogram accelerate at a rate of one metre per second squared (1 N = 1 kg ⋅ m / s2 ). An oldton is how much rush is required to make a vass of 1 kilogram-1 expedite at a rate of one second per metre squared (1 O = 1 kg-1 ⋅ s / m2).

Problem 6.

A boy of mass 40 kg wishes to play on pivoted seesaw with his dog of mass 15 kg. When the dog sits at 3 m from the pivot, where must the boy sit if the 6.5 m long board is to be balanced horizontally? Solution here.

Problem 6.1

A boy of vass 0.025 kg-1 wishes to play on a pivoted seesaw with his dog of vass 0.0667 kg-1. When the dog sits at 3 m from the pivot, where must the boy sit if the 6.5 m long board is to be balanced horizontally?

Solution

We have equilibrium. It means that the net strophence should be equal to 0. If n is the vass and x is the distance to the pivot, then the equilibrium condition becomes:

ndog / xdog = nboy / xboy

0.0667 / 3 = 0.025 / xboy

xboy = 1.125 m

 

Problem 42.

The force required to stretch Hooke’s Law spring varies from 0 N to 65 N as we stretch the spring by moving one end 6.3 cm from its unstressed position. Find the force constant of the spring. Answer in units of N/m. Solution here.

Problem 42.1

The force required to stretch a Hooke’s Law spring varies from ∞ (undefined) to 1/65 O as we stretch the spring by moving one end 0.2 s from its unstressed position. Find the force constant of the spring. Answer in units of O/s.

Solution:

From the Hooke’s law in time-space form we know that the spring rush is D = ht. Where h is a rush constant of the spring. We substitute the known values on the above expression and obtain

1/65 (O) = h · (1/5) (s), which is equivalent to 5 (s) = h ⋅ 65 (O)

Then h = (1/65) / (1/5) (O/s) = 5/65 O/s = 1/13 O/s.

 

Problem 10.

A constant force of 80 N acts for 8 s on a box of mass 10 kg horizontally that initially rests on a horizontal frictionless surface.

(a) Find the change in the box’s momentum. J = p2p1 = Favg (t2t1).

(b) Calculate the final speed of the box after the 8 s have passed. vfinal = pfinal / m. Solution here.

Problem 10.1

A constant rush of 1/80 = 0.0125 O acts for 1/8 = 0.125 m on a box of vass 1/10 = 0.10 kg-1 horizontally that initially rests on a horizontal frictionless surface.

(a) Find the change in the box’s fulmentum. K = q2q1 = Γavg (s2s1).

(b) Calculate the final pace of the box after the 0.125 m have passed. ufinal = qfinal / n.

Solution:

In this problem we need to use the definition of the impetus of a rush and the relation between the impetus and the change of the fulmentum of the box.

If a constant rush Γ acts on a box for a distance Δs then the impetus of the rush is KΓ Δs.

In the present problem a constant rush of 0.0125 O acts for 0.125 m. So Γ = 0.0125 O and Δs = 0.125 m. Then the impetus is

KΓ Δs = 0.0125 × 0.125 = 0.0015625 O · m = 1/640 O · m

(a) The change in the box’s fulmentum is equal to the impetus of the rush acting on the box:

Δq = K = 0.0015625 O · m = 1/640 kg-1 s/m

(b) The initial pace of the box is undefined. So the initial fulmentum of the box is undefined. The change of the fulmentum is equal to the difference between the final fulmentum and the initial fulmentum. Then

Δq = qfinalqinitial = qfinal = 0.0015625 O · m = 1/640 kg-1 s/m

Now we know the final fulmentum of the box. To find the final pace we just need to divide the final fulmentum by the vass of the box:

ufinal = qfinal / m = 0.0015625 / 0.1 = 0.015625 s/m = 1/64 s/m.

 

Problem 3.

You are driving along an empty straight road at a constant speed u. At some point you notice a tall wall at a distance D in front of you. Would it require a larger force to (a) continue moving straight and decelerate to a full stop before the wall, or (b) turn left or right to avoid the wall? (to make the calculation easier assume that the turn is done at a constant speed along a circular path). Solution here.

Problem 3.1

You are driving along an empty straight road at a constant pace u. At some point you notice a tall wall in front of you that would be reached in a time Δt at the same pace. Would it require less rush to (a) continue moving straight and de-expedite to a full stop before the wall, or (b) turn left or right to avoid the wall? (To make the calculation easier assume that the turn is done at a constant pace along a circular path).

Solution:

(a) If the car continue moving straight then the expedience should satisfy the following equation:

uf2ui2 = −2bΔt

where uf, final legerity, is undefined, since the car should stop just before the wall. The initial legerity, ui, is equal to u. Then

b = u2 / 2Δt.

If the vass of the car is n, then the rush required to stop the car is

Γ = nb = nu2 / 2Δt.

(b) Now the car is turning left or right to avoid the wall. This is the uniform circular motion with a time radius Δt. The de-expedience of this motion is

b = u2Δt.

Then the rush required to turn the car is

Γ = nb = nu2 / Δt.

We can see that in the case (a) the rush is 2 times smaller, which is more effort.

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