---

📘 FYBSc Internal Assignment Solutions (Explained for Class 11 Level)

---

Q1. A 0.150 kg particle moves along an x-axis according to:

$$x(t)=-13.00+2.00t+4.00t^2-3.00t^3$$

We have to find the net force on the particle at $t = 3.40 \, \text{s}$.

---

Step 1: Velocity is derivative of position

$$v(t) = \frac{dx}{dt} = 2.00 + (8.00t) - (9.00t^2)$$

Step 2: Acceleration is derivative of velocity

$$a(t) = \frac{dv}{dt} = 8.00 - 18.00t$$

Step 3: Force using Newton’s 2nd Law

$$F=m\cdot a$$

At $t = 3.40 \, \text{s}$:

$$a(3.40)=8.00-18.00(3.40)=8.00-61.20=-53.2{\text{m/s}}^2$$

Mass = 0.150 kg

$$F = 0.150 \times (-53.2) = -7.98 \, \text{N}$$

👉 Answer: Net Force = -7.98 N $\hat{i}$ (negative means direction is opposite x-axis)

---

Q2. A block of mass 8.5 kg on a 30° incline attached to a cord.

We need:
(a) Tension in cord
(b) Normal force on block
(c) Acceleration if cord is cut

---

Step 1: Forces acting on block

• Weight $W = mg = 8.5 \times 9.8 = 83.3 \, N$

• Component along incline = $W \sin \theta = 83.3 \times 0.5 = 41.65 \, N$

• Component perpendicular = $W \cos \theta = 83.3 \times 0.866 = 72.1 \, N$

(a) Tension: To balance down slope force

$$T = W \sin \theta = 41.65 \, N$$

(b) Normal force:

$$N = W \cos \theta = 72.1 \, N$$

(c) If cord is cut:
Only gravitational component acts → acceleration along incline

$$a=g\sin \theta =9.8\times 0.5=4.9m\mathrm{/}{s}^2$$

👉 Answer:
(a) Tension = 41.65 N
(b) Normal force = 72.1 N
(c) Acceleration = 4.9 m/s²

---

Q3. A block slides down a 60° incline with acceleration $g/2$. Find μk.

---

Step 1: Formula for acceleration with friction

$$a = g(\sin \theta - \mu_k \cos \theta)$$

Step 2: Plug values
Given $a = g/2$, $\theta = 60^\circ$, $\sin 60 = 0.866$, $\cos 60 = 0.5$

$$\frac{g}{2} = g(0.866 - \mu_k \times 0.5)$$

Divide by g:

$$0.5 = 0.866 - 0.5\mu_k$$
$$0.5\mu_k = 0.866 - 0.5 = 0.366$$
$${\mu }_k=\frac{0.366}{0.5}=0.732$$

👉 Answer: μk ≈ 0.73

---

Q4. Two blocks m and M connected via pulley. Find minimum & maximum M for no sliding.

---

Concept:

• For equilibrium, force due to hanging block $Mg$ should balance friction and keep system static.

• Range:

$$mg-{\mu }_{s}N\le Mg\le mg+{\mu }_{s}N$$

Here $N = mg$ (normal force). So:

$$(m - \mu_s m)g \leq Mg \leq (m + \mu_s m)g$$

Cancel g:

$$(m - \mu_s m) \leq M \leq (m + \mu_s m)$$

👉 Answer:

$$M_{min} = m(1 - \mu_s), \quad M_{max} = m(1 + \mu_s)$$

---

Q5. A bullet of mass 10 g and speed 250 m/s penetrates 5 cm into wood. Find force.

---

Step 1: Use work-energy principle
Work done by resistive force = Loss of kinetic energy

$$F \cdot d = \frac{1}{2} m v^2$$

Step 2: Insert values
Mass = 10 g = 0.01 kg
Distance = 5 cm = 0.05 m
Velocity = 250 m/s

$$F \times 0.05 = \frac{1}{2}(0.01)(250^2)$$ $$F \times 0.05 = 312.5$$ $$F = \frac{312.5}{0.05} = 6250 \, N$$

👉 Answer: Force = 6250 N

---

✅ Final Answers Summary

1. Net force at 3.40 s = -7.98 N (along -x direction)

2. Tension = 41.65 N, Normal force = 72.1 N, Acceleration (cord cut) = 4.9 m/s²

3. Coefficient of kinetic friction μk ≈ 0.73

4. Minimum M = m(1 - μs), Maximum M = m(1 + μs)

5. Force by tree on bullet = 6250 N

---