In our first example we will consider a very simple application of Newton\'s sec

Question

In our first example we will consider a very simple application of Newton's second law. A worker with spikes on his shoes pulls with a constant horizontal force of magnitude 20 N on a box with mass 40 kg resting on the flat, frictionless The box has no vertical acceleration, so the vertical components of the net force sum to zero. Nevertheless, for completeness, we show the vertical forces acting on the box.? The acceleration is given by Newton's second law. There is only one horizontal component of force, so we have The force is constant, so the acceleration is also constant. If we are given the initial position and velocity of the box, we can find the position and velocity at any later time from the equations of motion with constant acceleration. Two vertical (y) components of force act on the box-its weight w rightarrow and an upward normal force n exerted on the bottom of the box by the ice-but their vector sum is zero. If the box starts from rest and the worker pulls with a force of 50 N , what is the speed of the box after it has been pulled a distance of 0.36 m ? Express your answer in meters per second to two significant figures. REFLECT

Explanation / Answer

Given

Mass of the box = 40 kg

Initial velocity u = 0 m/s ( the box was at rest )

Applied force = 50 N

Distance travelled s = 0.36 m

Velocity after travelling distance s is v = ?

Solution

Here acceleration a = F/m

= 50 / 40

= 1.25 m/s2

For motion with constant acceleration

V2 = u2 + 2as

V2 = 0 + 2 x 1.25 x 0.36

V2 = 0.9

V = 0.9486 m/s

= 0.95 m/s (rounding off to two significant figures)

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.