Given is the following experimental setup to investigate Newton's second law.A cart carrying a mass M has a combined mass of m, is connected with a lightweight cord over a massless pulley to the mass m,.We measure the acceleration a as a function of the mass m,, where the mass m, is changed while keeping the total massmtota = m,+m, =0.8kg constant.Based on measurements results given in the table below,m, (kg)0.045 0.035 0.025 0.015ma0.649 0.467 0.395 0.213the equation of the least square line is given as y= 13.8 x+0.0171. Based on the equation of the least square line, calculate the experimental value for the gravitational acceleration gexo%3Ds22. Determine the percentage error (relative error) of the gravitational acceleration if the theoretical value is= 9.81m(Give the answer in 4 decimal places!)%error =%9 theoreticalFinish attempt

Given is the following experimental setup to investigate Newton's second law. A cart carrying a mass M has a combined mass of m, is connected with a lightweight cord over a massless pulley to the mass We measure the acceleration a as a function of the mass m2, where the mass m, is changed while keeping the total mass m, total = m,+m,=0.8kg constant. Based on measurements results given in the table below, m2 (kg) 0.045 0.035 0.025 0.015 m a 0.649 0.467 0.395 0.213 the equation of the least square line is given as y= 13.8 x+0.017.

Given is the following experimental setup to investigate Newton's second law. A cart carrying a mass M has a combined mass of m, is connected with a lightweight cord over a massless pulley to the mass We measure the acceleration a as a function of the mass m2, where the mass m, is changed while keeping the total mass m, total = m,+m,=0.8kg constant. Based on measurements results given in the table below, m2 (kg) 0.045 0.035 0.025 0.015 m a 0.649 0.467 0.395 0.213 the equation of the least square line is given as y= 13.8 x+0.017.

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter5: Newton's Law Of Motion

Section: Chapter Questions

Problem 30P: If the rocket sled shown in the previous problem starts with only one rocket burning, what is the...

See similar textbooks

Similar questions

Show that the acceleration of any object down an incline where friction behaves simply (that is, where fk=μkN)fk=μkN) is a=g(sinθ−μkcosθ).a=g(sinθ−μkcosθ). Note that the acceleration is independent of mass and reduces to the expression found in the previous problem when friction becomes negligibly small (μk=0).(μk=0).
Show that the acceleration of any object down an incline where friction behaves simply (that Is, where fk=kN ) is a=(sinkcos) . Note that the acceleration Is independent of mass and reduces to the expression found in the previous problem when friction becomes negligibly small (k=0) .
One paint bucket of known mass mA = 10.0kg is hanging by a massless cord from another paint bucket of known mass mB = 15.0kg, which is hanging by a massless cord. If the two buckets are pulled upward with a known acceleration of a = 2.0m/s2 by the upper cord, calculate the tension in cord 1 and the tension in cord 2 by (a) selecting appropriate common equations and making appropriate substitutions for the symbolic answer. (b) show the numeric substitution of given quantities for the final result. (c) Draw a free body diagram for each bucket.
Show that the acceleration of any object down an incline where friction behaves simply (that is, where fk = µkN ) is a = g(sinθ − µkcosθ). Note that the acceleration is independent of mass and reduces to the expression found in the previous problem when friction becomes negligibly small (µk = 0).
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…
On the moon, the acceleration due to gravity is equal to 1.67 m/s^2. How much does an astronaut weigh there, whose mass combined with his spacesuit is 171 kilograms.? Express your answer in units of Newtons.
In an experiment performed in a space station, a force of 60 N causes an object to have an acceleration equal to 4 m/s square. What is the object’s mass?
In procedure 2, you found the coefficient of kinetic friction by measuring the acceleration of a cart under the influence of the combined forces of friction, normal force, gravity, and tension in string from which masses were hanging. Suppose you add more masses to the hanging weight, therefore increasing the tension and hence the acceleration for this system to double (2 x) its original value. How many times of the old value of the coefficient of kinetic friction will the new value be?
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?
Consider two peoplepushing a toboggan with four children on it up a snowcovered slope. Construct a problem in which you calculatethe acceleration of the toboggan and its load. Include a freebody diagram of the appropriate system of interest as thebasis for your analysis. Show vector forces and theircomponents and explain the choice of coordinates. Amongthe things to be considered are the forces exerted by thosepushing, the angle of the slope, and the masses of thetoboggan and children.
If we want to find the size of the force necessary to just barely overcome static friction (in which case fs=μsN), we use the condition that the sum of the forces in both directions must be 0. Using some basic trigonometry, we can write this condition out for the forces in both the horizontal and vertical directions, respectively, as: Fcosθ−μsN=0 Fsinθ+N−mg=0 In order to find the magnitude of force F, we have to solve a system of two equations with both F and the normal force N unknown. Use the methods we have learned to find an expression for F in terms of m, g, θ, and μs (no N) What are the steps to solve this?
Someone (whose mass is 60 kg) is standing on a scale in an elevator that is accelerating upward, when he obtains a scale reading of 650 N, what is the acceleration of the elevator?