Lost on magnitude

  • Thread starter kimikims
  • Start date
  • #1
36
0
A person weighting 0.6kN rides in an elevator that has a downward acceleration of 1.4m/s^2. The acceleration of gravity is 9.8 m/s^2. What is the magnitude of the force of the elevator floor on the person? Answer in units of kN.

--

A 1317.9 kg car is coasting along a level road of 33 m/s. A constant breaking force is applied,such that the car is stopped in a distance of 63.7 m. What is the magnitude of the breaking force? Answer in units of N.

--

A 12000 kg sailboat experiences an eastward force 28300 N due to the tide pushing its hull while the wind pushes the sails with a force of 75900 N directed toward the northwest (45 degrees westward of North or 45 degrees northward of West). What is the magnitude of the resultant acceleration of the sailboat? Answer in units of m/s^2.

What is the direction of the boats acceleration? Answer in units of degrees (N of West).


---

I'm not sure what equations to use on these problems. Any help?
 

Answers and Replies

  • #2
there are only 2 equations
 
  • #3
36
0
Well what 2 equations are those? I don't know what to use to solve these problems.
 
  • #4
A person weighting 0.6kN rides in an elevator that has a downward acceleration of 1.4m/s^2. The acceleration of gravity is 9.8 m/s^2. What is the magnitude of the force of the elevator floor on the person? Answer in units of kN.

use F=ma

600 N=m 9.8 is how they arrived at the weight of a person. So the person weighed
600/9.8=m

the mass of the person is 61.2 kg

F=61.2kg*(9.8-1.4)

F=61.2kg *8.4 m/s sq
F=514.248 N
or 0.5 kN


The second one you need to a by using the formula V^2=V initial^2 + 2a(x1-x0)

0=V initial^2 +2a(63.7)

once you find a replace a in the formula F=ma
 
Last edited:
  • #5
36
0
In a "moment"um said:
A person weighting 0.6kN rides in an elevator that has a downward acceleration of 1.4m/s^2. The acceleration of gravity is 9.8 m/s^2. What is the magnitude of the force of the elevator floor on the person? Answer in units of kN.

use F=ma

600 N=m 9.8 is how they arrived at the weight of a person. So the person weighed
600/9.8=m

the mass of the person is 61.2 kg

F=61.2kg*(9.8-1.4)

F=61.2kg *8.4 m/s sq
F=514.248 N
or 0.5 kN


The second one you need to a by using the formula V^2=V initial^2 + 2a(x1-x0)

0=V initial^2 +2a(63.7)

once you find a replace a in the formula F=ma


0.5kN is wrong, is there a different way of solving this problem?
 
  • #6
Pyrrhus
Homework Helper
2,178
1
For the first problem use

Newton's 2nd Law

[tex] \sum_{i=1}^{n} \vec{F}_{i} = m \vec{a} [/tex]

You got movement on the y-axis, the forces are the normal and the weight, the acceleration is downward. Use a free body diagram. Momentum's answer algebraically is correct, if you use webassign or similar try combinations of figures.

For the third problem

Remember

[tex] F = ma [/tex]
 

Related Threads on Lost on magnitude

Replies
2
Views
9K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
3
Views
1K
Replies
6
Views
1K
Replies
4
Views
914
  • Last Post
Replies
10
Views
4K
Replies
1
Views
3K
Replies
3
Views
1K
Top