what is the largest speed this projectile can have without causing the cable to break

PHY131�������� ������������� Ch half dozen � 9 Test����������� Name

A force F = (2x i - 5y j) Newtons acts on an object every bit the object moves in the y dir from the origin to y = 4 meters.  Discover the work done on the object.

The x component is perpendicular; must omit (4/4) ; no work in the vertical direction

Work = ∫F∙dr (6/6)

Work = -5∫ydy

Work = -5 �y2 (14/xiv) ���� (0 to 4)

Piece of work = -40 Joules

A10.0 kg wood ball hangs from a 2.00 m-long wire. The maximum tension the wire tin can withstand without breaking is 500 N.�� A 1.00 kg projectile traveling horizontally hits and embeds itself in the wood brawl.What is the largest speed this projectile can have without causing the cable to break?

����������� (2/two) ���� (x/x) ���

FT ��� �� mg�� =�� mvf two /r

500 �11(x) = (10+1)fivef 2/2

780/11 = 5f 2

pinit ��� =��� pfinal ��� (ten/ten) ����

mp vp = (mp+mb )5f

(1)5p �� = (i + 10)vf

5f ������� = vp / xi

(3/3) ���� 780/11 = fivep two / 112

5p = 92.six m/s

3 blocks are connected by a massless string that passes over frictionless pulleys as shown.  The 60� incline is frictionless but the table has a frictional coef of 0.2.  Find the tension at signal �A� in the string.

����������� (5/5) ��� (5/5) ������ (five/5) ���� (4/4) ����

��������� FNet ������������ ����� Ff ��� =�� mT ���� a ���

[iii(k)sin60� - ane(g)] � [�(2)g] = (3+1+two) a

a = ii k/due south2

FT = grand (gsinθ � a) �� �� (6/6) ����

FT = 20 North

You identify your water bottle on acme of your apartment dashboard.  The coefficient of static friction between your water bottle and the panel is μs = 0.300.  How fast can yous drive on a horizontal roadway around a gradual correct turn of radius 40.0 grand earlier the bottle slides?

Ff = μFN

Fc = mvii/r

Logic: The frictional force provides the centripetal forcefulness towards the middle

Ff = Fc

μFNorth = mvii/r

(14/xiv) ��� (10/10) ����

μmg = mv2/r

v2 =  μ    g    r

v2 = 0.5(x)(lxxx)

v = xx m/s

(1/one) ����

PHY131�������� ������������� Ch 6 � 9 Exam����������� Proper name

An odd jump with a forcefulness of F = -grand x2 is compressed from three cm to i cm.What was the free energy required to compress the spring if k = 100 N/cm2?

Work =�� F�� dx ������� (6/6) ����

Work = -k ∫ten2 dx ��� (14/14) ����

Work = k/iii (x0 3 -xf 3)���� (from 3 to 1)

Piece of work = 800/3 N●cm �� (5/five) ����

A brawl is launched upwardly a semicircular chute in such a way that at the summit of the chute, only before information technology goes into free fall, the brawl has a centripetal acceleration of magnitude 3g.How far from the bottom of the chute does the ball land?

(8/8) ����

m3g = mv2/R

vx 2 = (3Rg)

(8/8) ����

y�� = � g t2

2R = � g t2

t2 = (4R/g)

fivex =��� Δx �� / Δt ���� (8/eight) ����

Δx =����� v10 �������� t

Δx = (3Rg) (4R/g)

Δx = R(12) ��� (i/i) ����

An elastic cord has a force of F = 8t + 4 Newtons .What is the change of velocity of a 1 kg object if the launch time is 0.5 seconds.

���� F��� dt ��� �����= k dv �������������� (6/6) ����

8∫tdt + 4∫dt= m��� Δv (14/14) ����

iv t2 + 4t���� �����= m��� Δv

i N + 2 N�� ������= 1 kg Δv ���������� (4/4) ����

Δv = iii one thousand/s������������������������������� (ane/ane) ����

Two blocks both of mass 10 kg are connected by a massless string that passes over a frictionless pulley.  The 60� and 30� inclines both have a frictional coef of 0.1.  Find the tension in the string.

Fin dir = sin lx� mg

Fin dir = 0.866(100N)

Fin dir-left = 86.6 N

FT-lx� =yard��    a����� - Ff

FT-60� = ten(sin60�grand-a) - 5

FT-sixty� = 70.one N

FNet -���������������������� Ff����������� = chiliadfull   a

(86.6 � 50) - (.1*cos60mg +.one*sin60mg) = (ten+10) a

(4/4) �� (4/4) (vii/7) ���� (ii/two) ���� ��������� (2/two) ���� ������ (4/4)

36.6���������� � ( v��������������� +�� 8.66�� )���� = 20 a

a = ane.15 1000/s2 (1/1) ����

Fin dir = sin 30� mg

Fin dir = � (100 North)

Fin dir-right = 50 N

FT-thirty� =m��� a������� -�� Ff

FT-30� = 10 (sin30�g+a) + 8.66

FT-30� = 70.1 Northward

PHY131�������� ������������� Ch 6 � nine Exam����������� Name

A 4 kg object is moving in a aeroplane with its y-coordinate given past y = (2tthree + 8t +ane) meters.  Notice the magnitude of the velocity and the net force acting on this object at t = iii sec.

y = 2t3 + 8t + 1

vy = 6t2 + 8 v = dx / dt

ay = 12t����� a = dv / dt

vy = 6t2 + 8

5y = vi*iii2 + 8

vy = 62 m/s

Fy = m ay

Fy = 4 (12t)

Fy = 144 North

Ii blocks both of mass 10 kg are connected by a massless string that passes over a frictionless pulley.  The 60� and xxx� inclines are frictionless.  Detect the tension in the cord.

sin threescore mg

Fin dir = 0.866(100N)

Fin dir-left = 86.half dozen Northward

FT-sixty� ������� = m�� a

FT-60� ������� = 10 (sin 60 g - a) �������

FT-60� ������� = x (seven.82) ������

FT-lx� ������� = 68.3 N

Fin dir = sin 30 mg

Fin dir = � (100 Northward)

Fin dir-correct = l North

FT-30� ������� = m����� a

FT-30� ������� = x (sin thirty g + a)

FT-thirty� ������� = 10 (half-dozen.83) ������

FT-30� ������� = 68.iii N

FInternet ����������������� =�� thoutotal a

(86.650) ����� = (ten+10) a

a = 1.83 m/s2 ��

Two basketballs are on a collision course.  1 is big sized (150 g) and the other is a junior ball of 100 grams.  The large brawl is traveling 2 m/s; the pocket-size ball is traveling at -four k/s.  What are their final velocities?

Initial:v1f - 52f = half-dozen m/south������������

Final:v2f - v1f = six one thousand/south;���� v2f = 6 m/southward + 51f

m151 ����� + 1000iiv2 ����������� = k1f 51f ����������� + ��������� thousand2f five2f

0.15(2) + 0.one(-4) ���������� = 0.15 v1f ���������� + ��������� 0.3 v2f

0.3 ����� - 0.4 ��������������� = 0.15v1f ���������� + 0.3(vi + v1f )

-0.1������������������������������ = 0.15v1f ���������� + i.8 + 0.3v1f

v1f ������� = -four.22 m/s������������������ �����������

v2f ������ = 2.44 m/due south

Mass m1 on the frictionless table and is connected by a string through a hole in the tabular array to a hanging mass yardii.With what speed must mane rotate in a circle of radius r if m2 is to remain hanging at rest?

m2 g = m1 v2 / r

v = (m2 g r / 1000one)

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Source: https://www.cpp.edu/~skboddeker/131/quiz/09-4fall-e3.htm

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