Question

If an object has a mass of 35 kilograms and accelerates at 10 m/s2, what was the force that pushed the object?

Answer 1

Provided that:

• Mass = 35 kilograms
• Acceleration = 10 m/s²

To calculate:

• Force

Solution:

• Force = 350 N

Using concept:

• Newton's second law of motion that is the formula of force

Using formula:

• F = ma

(Where, F denotes force, m denotes mass and a denotes acceleration)

Required solution:

→ F = ma

→ F = 35(10)

→ F = 35 × 10

→ F = 350

→ Force = 350 Newton

Additional information:

$\begin{gathered}\boxed{\begin{array}{c}\\ \bf What \: is \: acceleration? \\ \\ \sf The \: rate \: of \: change \: of \: velocity \: of \: an \\ \sf object \: with \: respect \: to \: time \\ \sf is \: known \: as \: acceleration. \\ \\ \sf \star \: Negative \: acceleration is \: known \: as \: deacceleration. \\ \sf \star \: Deacceleration \: is \: known \: as \: retardation. \\ \sf \star \: It's \: SI \: unit \: is \: ms^{-2} \: or \: m/s^2 \\ \sf \star \: It \: may \: be \: \pm ve \: or \: 0 \: too \\ \sf \star \: It \: is \: a \: vector \: quantity \\ \\ \bf Conditions \: of \pm ve \: or \: 0 \: acceleration \\ \\ \sf \odot \: Positive \: acceleration: \: \sf When \: \bf{u} \: \sf is \: lower \: than \: \bf{v} \\ \sf \odot \: Negative \: acceleration: \: \sf When \: \bf{v} \: \sf is \: lower \: than \: \bf{u} \\ \sf \odot \: Zero \: acceleration: \: \sf When \: \bf{v} \: \sf and \: \bf{u} \: \sf are \: equal \end{array}}\end{gathered}$