In Newton's second law experiment, the time (t) needed to cover a fixed distance (S= 0.800 m) is studied as function of the total mass of the system (M m1+ m2) at a constant hanging mass (m2). The time (t) is related to (MT) by 2SM t2 = m Where g: acceleration due to gravity = 9.81 m/s2 t (s?) Slope = 1.63 MT (kg) The value of (m2) in grams is: 1.110 O2.0.27 3.100 4.90 5.0.22

In Newton's second law experiment, the time (t) needed to cover a fixed distance (S= 0.800 m) is studied as function of the total mass of the system (M m1+ m2) at a constant hanging mass (m2). The time (t) is related to (MT) by 2SM t2 = m Where g: acceleration due to gravity = 9.81 m/s2 t (s?) Slope = 1.63 MT (kg) The value of (m2) in grams is: 1.110 O2.0.27 3.100 4.90 5.0.22

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter4: The Laws Of Motion

Section: Chapter Questions

Problem 13P

See similar textbooks

Similar questions

A particle of mass 1.05 kg is subject to a force that is always pointed towards the East but whose magnitude changes linearly with time t. The magnitude of the force is given as F = 4t, and has units of newtons. Let the x-axis point towards the East. Determine the change in x-coordinate in meters of the particle Δx between t = 0 and t = 2.4 if the initial velocity is 17.5 m/s, and pointed in the same direction as the force.
In Fig, a slab of mass m1 =40 kg rests on a frictionless floor, and a block of mass m2 =10kg rests on top of the slab. Between block and slab, the coefficient of static friction is 0.60, and the coefficient of kinetic friction is 0.40. A horizontal force of magnitude 100 N begins to pull directly on the block, as shown. In unit-vector notation, what are the resulting accelerations of (a) the block and (b) the slab?
Consider the 63 kg ice skater being pushed by two others shown in the figure. The coefficient of static friction is μs=0.4 and kinetic is μk=0.02.Randomized Variablesm = 63 kgF1 = 238 NF2 = 175 N Find the magnitude of Ftot, the total force exerted on her by the others, given that the magnitudes F1 and F2 are 238 N and 175 N, respectively in Newtons. Find the direction of Ftot (in degrees relative to the horizontal), the total force exerted on her by the others, given that the magnitudes F1 and F2 are 238 N and 175 N, respectively .What is the maximum value of the static friction force, in Newtons, that can act on the skater before she moves? What is her acceleration assuming she is already moving in the direction of Ftot in m/s2?
In the friction experiment, if the friction block has a mass of 599 g, and it started to move at a constant speed after giving it a slight push when the mass in the hanger is 284 g (including the hanger), what is the coefficient of kinetic friction?
According to a simplified model of a mammalian heart, at each pulse, approximately 20 g of blood is accelerated from 0.26 m/s to 0.35 m/s during a period of 0.10 s. What is the magnitude of the force exerted by the heart muscle? Express your answer to two significant figures and include the appropriate units. F= ______ units I got 0.02N but it is wrong.
Suppose that a projectile of mass m moves in a verticalplane in the atmosphere near the surface of the earth underthe influence of two forces: a downward gravitationalforce of magnitude mg, and a resistive force FR that isdirected opposite to the velocity vector v and has magnitudek v2 (where v = |v| is the speed of the projectile;see Fig. 4.1.15). Show that the equations of motion of theprojectile are mx'' = -kvx' , my'' = -kvy' -mg,
Consider 2 forces acting on a mass of 2.42 kg. F1 = 10.4 N @ 18.1 degrees and F2 = 7.07 N @ 66.4 degrees where the angles are measured relative to the +x direction. What is the magnitude of the acceleration produced considering only these 2 forces acting?
Consider 2 forces acting on a mass of 2.91 kg. F1 = 9.3 N @ 22 degrees and F2 = 6.28 N @ 83.7 degrees where the angles are measured relative to the +x direction. What is the magnitude of the acceleration produced considering only these 2 forces acting? everything in 3 sig figs
A 3.5 kg particle moves along an x axis according to x(t) = −12 + 5t2 − 4t3, with xin meters and t in seconds. In unit-vector notation, what is the net force acting on theparticle at t = 6 s?
A particle of mass 1.3 kg is subject to a force that is always pointed towards the East but whose magnitude changes linearly with time t. The magnitude of the force is given as F = 5t, and has units of newtons. Let the x-axis point towards the East. a)Determine the change in the velocity Δv, in meters per second, of the particle between t = 0 and t = 1.4 sec. b)
Recall that Newton’s Equation of Motion for a Body of Mass, m, is F = m(dv/dt). Assume the Initial Velocity of the Projectile is v0, and the Gravitational Acceleration Constant is g, where the Gravitational Force is Fg = -mg. Note that the Maximum Height will Occur at a Time, tm, when the Velocity Goes to Zero, v(tm) = 0. a) Simply Solve the Equation of Motion for the Velocity as a Function of Time, and Determine the Time it Takes the Projectile to Reach the Maximum Height, tm, As a Function of the Given Parameters. b) Include a Force of Air Resistance that is Proportional to the Velocity of the Form Fr = -kmv. Again, Solve the Equation of Motion for the Velocity as a Function of Time, by Direct Integration or by Choosing an Appropriate Ansatz Form for the Solution. Again, Determine an Expression for the Time it Takes the Projectile to Reach the Maximum Height, tm, As a Function of the Given Parameters. c) Finally, Compare the Two Results by First Checking that they Agree for a Small…
A block of mass M with a contact area A slides down an inclined plane with a coefficient of friction of 0.4 and travels a distance L in time T. How long would it take another block with the same mass and composition , but with contact area 2A to slide at the same distance L?