Physics Lab: Newton's Laws & SUVAT Kinematics

High School Physics • Perpendicular Forces & Kinematics ($g = 9.8\text{ m/s}^2$)

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💡 Quick Formula Reference: Newton's Laws & SUVAT Kinematics (Click to expand) Standard: $g = 9.8\text{ m/s}^2$
Newton's Laws & Forces:

Static Equilibrium: $\sum F_x = 0$, $\sum F_y = 0$

Newton's 2nd Law: $\sum F = m \cdot a$

Weight Force: $W = m \cdot g$ (acting downwards)

SUVAT Kinematic Equations:

$v = u + at \implies a = \frac{v - u}{t}$

$s = ut + \frac{1}{2}at^2 \implies a = \frac{2(s - ut)}{t^2}$

$v^2 = u^2 + 2as \implies a = \frac{v^2 - u^2}{2s}$

Variables & SI Units:

$u$: Initial Velocity ($\text{m/s}$), $v$: Final Velocity ($\text{m/s}$)

$a$: Acceleration ($\text{m/s}^2$), $t$: Time elapsed ($\text{s}$)

$s$: Displacement ($\text{m}$), $F, W, T$: Force ($\text{N}$), $m$: Mass ($\text{kg}$)

Step-by-Step Problem Solving Framework

1) Free Body Diagram (FBD): Toggle only the strictly perpendicular forces acting along the respective axis ($X$ or $Y$). No diagonal forces are present. 2) Kinematics (SUVAT): If the system is not in equilibrium, compute the acceleration $a$ from the velocity, displacement, or time given. If in static equilibrium, $a = 0$. 3) Force Equation: Formulate $\sum F = 0$ or $\sum F = m \cdot a$ and solve for the target force in Newtons ($\text{N}$).

1 Static Equilibrium • Horizontal X-Axis
Pending

Stationary Crate Pulled by Two Ropes

In a school warehouse, a heavy storage crate of mass $m = 25\text{ kg}$ sits completely at rest on a frictionless horizontal floor. Student A pulls horizontally to the right with a constant force $F_1 = 80\text{ N}$. Student B holds the crate from moving by pulling horizontally to the left with an unknown force $F_2$.

Given: $m = 25\text{ kg}$, $F_1 = 80\text{ N}$ (to the right), $a_x = 0\text{ m/s}^2$ (at rest).
Step 1: Construct the FBD (Horizontal Forces)

Click buttons below to toggle vector arrows onto the crate. Select all forces acting on the crate:

Because the crate remains stationary ($a_x = 0$), which statement correctly expresses Newton's 1st Law?

Calculate the magnitude of force $F_2$ required to maintain static equilibrium:

$F_2 =$ $\text{N}$
2 Static Equilibrium • Vertical Y-Axis
Pending

Backpack Resting on a Lab Table

A student's backpack with mass $m = 6.5\text{ kg}$ rests motionless on a horizontal laboratory table. There are no cables or horizontal pushes applied. Using the local gravitational acceleration $g = 9.8\text{ m/s}^2$, determine the downward weight $W$ and the upward contact normal force $F_N$ exerted by the table.

Given: $m = 6.5\text{ kg}$, $g = 9.8\text{ m/s}^2$, $a_y = 0\text{ m/s}^2$ (vertical static equilibrium).
Step 1: Construct Vertical FBD (Y-Axis)

Activate the vertical perpendicular forces present for this stationary backpack:

Substitute $m = 6.5\text{ kg}$ and $g = 9.8\text{ m/s}^2$:

$W = 6.5 \cdot 9.8 =$ $\text{N}$

Apply equilibrium $\sum F_y = F_N - W = 0 \implies F_N = W$:

$F_N =$ $\text{N}$
3 Horizontal Dynamics (X) • SUVAT Link ($v = u + at$)
Pending

Accelerated Laboratory Air Track Glider

A glider of mass $m = 1.2\text{ kg}$ starts from rest ($u = 0\text{ m/s}$) on a horizontal, frictionless linear air track. A single constant horizontal net force $F$ pushes it towards the right. After a time of $t = 3.0\text{ s}$, the glider reaches a final speed $v = 4.5\text{ m/s}$.

Given: $m = 1.2\text{ kg}$, $u = 0\text{ m/s}$, $v = 4.5\text{ m/s}$, $t = 3.0\text{ s}$.
Step 1: Construct FBD (Single Net Force in X)

Include the horizontal pushing force and vertical support forces (no friction):

Use $v = u + at \implies a = \frac{v - u}{t}$:

$a = \frac{4.5 - 0}{3.0} =$ $\text{m/s}^2$

Multiply the mass $m = 1.2\text{ kg}$ by the computed acceleration:

$F = 1.2 \cdot a =$ $\text{N}$
4 Vertical Dynamics (Y) • SUVAT Link ($s = ut + \frac{1}{2}at^2$)
Pending

Crane Cable Hoisting a Steel Container

A construction container of mass $m = 40\text{ kg}$ is hoisted vertically upwards from the ground starting from rest ($u = 0\text{ m/s}$). The crane cable exerts an upward vertical tension force $T$. The container ascends a vertical distance $s = 6.0\text{ m}$ in $t = 2.0\text{ s}$ with uniform acceleration.

Given: $m = 40\text{ kg}$, $u = 0\text{ m/s}$, $s = 6.0\text{ m}$, $t = 2.0\text{ s}$, $g = 9.8\text{ m/s}^2$.
Step 1: Construct Vertical FBD (Hoisted Body)

Select the vertical forces acting on the suspended container:

Since $u=0$, solve for acceleration $a = \frac{2s}{t^2} = \frac{2(6.0)}{2.0^2}$:

$a =$ $\text{m/s}^2$

Equation: $\sum F_y = T - W = m \cdot a \implies T = W + m \cdot a$

Note: $W = m \cdot g = 40 \cdot 9.8 = 392\text{ N}$.

$T =$ $\text{N}$
5 Horizontal Braking (X) • SUVAT Link ($v^2 = u^2 + 2as$)
Pending

Emergency Braking of a Test Vehicle

A test sled of mass $m = 500\text{ kg}$ is cruising initially towards the right at a velocity $u = 20\text{ m/s}$. The emergency brakes engage, producing a constant retarding braking force that brings the vehicle to a full stop ($v = 0\text{ m/s}$) over a straight distance of $s = 40\text{ m}$.

Given: $m = 500\text{ kg}$, $u = 20\text{ m/s}$, $v = 0\text{ m/s}$, $s = 40\text{ m}$.
Step 1: Construct Braking FBD

The car moves to the right, but is slowing down. In which direction must the net braking force point?

Solve for $a = \frac{v^2 - u^2}{2s} = \frac{0^2 - 20^2}{2 \cdot 40} = \frac{-400}{80}$:

$a =$ $\text{m/s}^2$

Calculate the absolute magnitude (positive value) of the braking force:

$F_{\text{brake}} =$ $\text{N}$

Workshop Assessment & Summary

Verify all 5 problems at once to check your mastery of Newton's Laws connected with SUVAT kinematics, or print your worksheet to submit.