please I need help and detailed explanations to solving these questions..




1. A 300g mass hangs at the end of a string. A second string hangs from the bottom of that mass and supports a 900g mass. (a) Find the tension in each string when the masses are accelerating upward at 0.700m/s2. (b) Find the tension in each string when the masses are accelerating downward at 0.700m/s2.



2.
A horizontal force of 200N is required to cause a 15kg block to slide up a 20° incline with an acceleration of 25cm/s2. Find (a) the friction force on the block and (b) the coefficient of friction.


3.
A lawn tennis ball of mass m and speed v strikes a wall perpendicularly and rebounds with undiminished speed. If the time of collision is t, what is the average force exerted by the ball on the wall?


4. A block of mass 2kg is pulled along a smooth horizontal surface by a horizontal force P. What is the value of the normal force exerted on the block by the surface, and determine the force P required to give the block horizontal velocity of 4ms-1 in 2s

How much will you pay?

Let me tackle it once and for all....

Firstly, I'm not a guru
Secondly, I may be wrong.

I'll attempt 3 and 4

For question 3, according Newton's second law.

F = change in momentum/time
From the question, neither mass nor velocity changes...

Hence F = Mv - Mv/t
F = 0/t
F= 0... i.e no force was applied

for question 4, although this is quite confusing;

a. the force exerted by the surface should be the Normal Reaction.
Hence, R = mg
R = 2 * 10(provided g=10m/s); R= 20N

b. Using Newtons second law
F = M(v - u)/t.... Though it doesn't state the block was at rest. Assuming it was

F = 2(4 - 0)/2;
F= 4N

BjDreamchaser:

1. A 300g mass hangs at the end of a string. A second string hangs from the bottom of that mass and supports a 900g mass. (a) Find the tension in each string when the masses are accelerating upward at 0.700m/s2. (b) Find the tension in each string when the masses are accelerating downward at 0.700m/s2.

Soln
M¹=300/1000= 0.3kg
M²=900/1000= 0.9kg
a=0.700m/s²
g=10m/s²

(a)
Tension = mg + ma (upward acceleration)
=0.3*10 + 0.3*0.700
=3 + 0.21
=3.21N

Tension in second string =>
0.9*10 + 0.9*0.700
9+0.63
=9.63
Since the second mass hangs on M¹ the total force/tension in string one will be
3.21N + 9.63 (the extra force pulling M¹)
=12.84N

---

(b)
Tension with downward acceleration = mg - ma
0.3*10 - 0.3*0.700
3 - 0.21
=2.79N

Second string
0.9*10 - 0.9*0.700
=9 - 0.63
=8.37N

Total Tension in First string with downward acceleration. . =8.37 + 2.79
=11.08N

Optimist1998:

Soln
M¹=300/1000= 0.3kg
M²=900/1000= 0.9kg
a=0.700m/s²
g=10m/s²

(a)
Tension = mg + ma (upward acceleration)
=0.3*10 + 0.3*0.700
=3 + 0.21
=3.21N

Tension in second string =>
0.9*10 + 0.9*0.700
9+0.63
=9.63
Since the second mass hangs on M¹ the total force/tension in string one will be
3.21N + 9.63 (the extra force pulling M¹)
=12.84N

---

(b)
Tension with downward acceleration = mg - ma
0.3*10 - 0.3*0.700
3 - 0.21
=2.79N

Second string
0.9*10 - 0.9*0.700
=9 - 0.63
=8.37N

Total Tension in First string with downward acceleration. . =8.37 + 2.79
=11.08N

100% right. Thank you boss...